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14.2: Limits and Continuity


Learning Objectives

  • Calculate the limit of a function of two variables.
  • Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach.
  • State the conditions for continuity of a function of two variables.
  • Verify the continuity of a function of two variables at a point.
  • Calculate the limit of a function of three or more variables and verify the continuity of the function at a point.

We have now examined functions of more than one variable and seen how to graph them. In this section, we see how to take the limit of a function of more than one variable, and what it means for a function of more than one variable to be continuous at a point in its domain. It turns out these concepts have aspects that just don’t occur with functions of one variable.

Limit of a Function of Two Variables

Recall from Section 2.5 that the definition of a limit of a function of one variable:

Let (f(x)) be defined for all (x≠a) in an open interval containing (a). Let (L) be a real number. Then

[lim_{x→a}f(x)=L]

if for every (ε>0,) there exists a (δ>0), such that if (0<|x−a|<δ) for all (x) in the domain of (f), then

[|f(x)−L|<ε.]

Before we can adapt this definition to define a limit of a function of two variables, we first need to see how to extend the idea of an open interval in one variable to an open interval in two variables.

Definition: (delta) Disks

Consider a point ((a,b)∈mathbb{R}^2.) A (δ) disk centered at point ((a,b)) is defined to be an open disk of radius (δ) centered at point ((a,b)) —that is,

[{(x,y)∈mathbb{R}^2∣(x−a)^2+(y−b)^2<δ^2}]

as shown in Figure (PageIndex{1}).

The idea of a (δ) disk appears in the definition of the limit of a function of two variables. If (δ) is small, then all the points ((x,y)) in the (δ) disk are close to ((a,b)). This is completely analogous to x being close to a in the definition of a limit of a function of one variable. In one dimension, we express this restriction as

[a−δ

In more than one dimension, we use a (δ) disk.

Definition: limit of a function of two variables

Let (f) be a function of two variables, (x) and (y). The limit of (f(x,y)) as ((x,y)) approaches ((a,b)) is (L), written

[lim_{(x,y)→(a,b)}f(x,y)=L]

if for each (ε>0) there exists a small enough (δ>0) such that for all points ((x,y)) in a (δ) disk around ((a,b)), except possibly for ((a,b)) itself, the value of (f(x,y)) is no more than (ε) away from (L) (Figure (PageIndex{2})).

Using symbols, we write the following: For any (ε>0), there exists a number (δ>0) such that

[|f(x,y)−L|<ε]

whenever

[0

Proving that a limit exists using the definition of a limit of a function of two variables can be challenging. Instead, we use the following theorem, which gives us shortcuts to finding limits. The formulas in this theorem are an extension of the formulas in the limit laws theorem in The Limit Laws.

Limit laws for functions of two variables

Let (f(x,y)) and (g(x,y)) be defined for all ((x,y)≠(a,b)) in a neighborhood around ((a,b)), and assume the neighborhood is contained completely inside the domain of (f). Assume that (L) and (M) are real numbers such that

[lim_{(x,y)→(a,b)}f(x,y)=L]

and

[lim_{(x,y)→(a,b)}g(x,y)=M,]

and let (c) be a constant. Then each of the following statements holds:

Constant Law:

[lim_{(x,y)→(a,b)}c=c]

Identity Laws:

[lim_{(x,y)→(a,b)}x=a]

[lim_{(x,y)→(a,b)}y=b]

Sum Law:

[lim_{(x,y)→(a,b)}(f(x,y)+g(x,y))=L+M]

Difference Law:

[lim_{(x,y)→(a,b)}(f(x,y)−g(x,y))=L−M]

Constant Multiple Law:

[lim_{(x,y)→(a,b)}(cf(x,y))=cL]

Product Law:

[lim_{(x,y)→(a,b)}(f(x,y)g(x,y))=LM]

Quotient Law:

[lim_{(x,y)→(a,b)}dfrac{f(x,y)}{g(x,y)}=dfrac{L}{M} ext{ for } M≠0]

Power Law:

[lim_{(x,y)→(a,b)}(f(x,y))^n=L^n]

for any positive integer (n).

Root Law:

[lim_{(x,y)→(a,b)}sqrt[n]{f(x,y)}=sqrt[n]{L}]

for all (L) if (n) is odd and positive, and for (L≥0) if n is even and positive.

The proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws to finding limits of various functions.

Example (PageIndex{1}): Finding the Limit of a Function of Two Variables

Find each of the following limits:

  1. (displaystyle lim_{(x,y)→(2,−1)}(x^2−2xy+3y^2−4x+3y−6))
  2. (displaystyle lim_{(x,y)→(2,−1)}dfrac{2x+3y}{4x−3y})

Solution

a. First use the sum and difference laws to separate the terms:

[egin{align*} lim_{(x,y)→(2,−1)}(x^2−2xy+3y^2−4x+3y−6) = left(lim_{(x,y)→(2,−1)}x^2 ight)− left(lim_{(x,y)→(2,−1)}2xy ight)+ left(lim_{(x,y)→(2,−1)}3y^2 ight)−left(lim_{(x,y)→(2,−1)}4x ight) + left(lim_{(x,y)→(2,−1)}3y ight)−left(lim_{(x,y)→(2,−1)}6 ight). end{align*}]

Next, use the constant multiple law on the second, third, fourth, and fifth limits:

[egin{align*} =(lim_{(x,y)→(2,−1)}x^2)−2(lim_{(x,y)→(2,−1)}xy)+3(lim_{(x,y)→(2,−1)}y^2)−4(lim_{(x,y)→(2,−1)}x) [4pt] +3(lim_{(x,y)→(2,−1)}y)−lim_{(x,y)→(2,−1)}6.end{align*}]

Now, use the power law on the first and third limits, and the product law on the second limit:

[egin{align*} left(lim_{(x,y)→(2,−1)}x ight)^2−2left(lim_{(x,y)→(2,−1)}x ight) left(lim_{(x,y)→(2,−1)}y ight)+3left(lim_{(x,y)→(2,−1)}y ight)^2 −4left(lim_{(x,y)→(2,−1)}x ight)+3left(lim_{(x,y)→(2,−1)}y ight)−lim_{(x,y)→(2,−1)}6. end{align*}]

Last, use the identity laws on the first six limits and the constant law on the last limit:

[egin{align*} lim_{(x,y)→(2,−1)}(x^2−2xy+3y^2−4x+3y−6) = (2)^2−2(2)(−1)+3(−1)^2−4(2)+3(−1)−6 [4pt] =−6. end{align*}]

b. Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law,

[egin{align*} lim_{(x,y)→(2,−1)}(4x−3y) =lim_{(x,y)→(2,−1)}4x−lim_{(x,y)→(2,−1)}3y [4pt] =4(lim_{(x,y)→(2,−1)}x)−3(lim_{(x,y)→(2,−1)}y) [4pt] =4(2)−3(−1)=11. end{align*}]

Since the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law:

[egin{align*} lim_{(x,y)→(2,−1)}(2x+3y) =lim_{(x,y)→(2,−1)}2x+lim_{(x,y)→(2,−1)}3y [4pt] =2(lim_{(x,y)→(2,−1)}x)+3(lim_{(x,y)→(2,−1)}y) [4pt] =2(2)+3(−1)=1. end{align*}]

Therefore, according to the quotient law we have

[egin{align*} lim_{(x,y)→(2,−1)}dfrac{2x+3y}{4x−3y} =dfrac{displaystyle lim_{(x,y)→(2,−1)}(2x+3y)}{displaystyle lim_{(x,y)→(2,−1)}(4x−3y)} [4pt] =dfrac{1}{11}. end{align*}]

Exercise (PageIndex{1}):

Evaluate the following limit:

[lim_{(x,y)→(5,−2)}sqrt[3]{dfrac{x^2−y}{y^2+x−1}}. onumber]

Hint

Use the limit laws.

Answer

[displaystyle lim_{(x,y)→(5,−2)}sqrt[3]{dfrac{x^2−y}{y^2+x−1}}=dfrac{3}{2} onumber]

Since we are taking the limit of a function of two variables, the point ((a,b)) is in (mathbb{R}^2), and it is possible to approach this point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the path taken toward ((a,b)). If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken.

Example (PageIndex{2}): Limits That Fail to Exist

Show that neither of the following limits exist:

  1. (displaystyle lim_{(x,y)→(0,0)}dfrac{2xy}{3x^2+y^2})
  2. (displaystyle lim_{(x,y)→(0,0)}dfrac{4xy^2}{x^2+3y^4})

Solution

a. The domain of the function (f(x,y)=dfrac{2xy}{3x^2+y^2}) consists of all points in the (xy)-plane except for the point ((0,0)) (Figure (PageIndex{3})). To show that the limit does not exist as ((x,y)) approaches ((0,0)), we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point ((0,0)). First, consider the line (y=0) in the (xy-plane.) Substituting (y=0) into (f(x,y)) gives

[f(x,0)=dfrac{2x(0)}{3x^2+0^2}=0 onumber]

for any value of (x). Therefore the value of (f) remains constant for any point on the (x)-axis, and as (y) approaches zero, the function remains fixed at zero.

Next, consider the line (y=x). Substituting (y=x) into (f(x,y)) gives

[f(x,x)=dfrac{2x(x)}{3x^2+x^2}=dfrac{2x^2}{4x^2}= frac{1}{2}. onumber]

This is true for any point on the line (y=x). If we let (x) approach zero while staying on this line, the value of the function remains fixed at ( frac{1}{2}), regardless of how small (x) is.

Choose a value for ε that is less than (1/2)—say, (1/4). Then, no matter how small a (δ) disk we draw around ((0,0)), the values of (f(x,y)) for points inside that (δ) disk will include both (0) and ( frac{1}{2}). Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.

b. In a similar fashion to a., we can approach the origin along any straight line passing through the origin. If we try the (x)-axis (i.e., (y=0)), then the function remains fixed at zero. The same is true for the (y)-axis. Suppose we approach the origin along a straight line of slope (k). The equation of this line is (y=kx). Then the limit becomes

[egin{align*} lim_{(x,y)→(0,0)}dfrac{4xy^2}{x^2+3y^4} = lim_{(x,y)→(0,0)}dfrac{4x(kx)^2}{x^2+3(kx)^4} = lim_{(x,y)→(0,0)}dfrac{4k^2x^3}{x^2+3k^4x^4} =lim_{(x,y)→(0,0)}dfrac{4k^2x}{1+3k^4x^2} = dfrac{displaystyle lim_{(x,y)→(0,0)}(4k^2x)}{displaystyle lim_{(x,y)→(0,0)}(1+3k^4x^2)} = 0. end{align*}]

regardless of the value of (k). It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation (x=y^2). Substituting (y^2) in place of (x) in (f(x,y)) gives

[egin{align*}lim_{(x,y)→(0,0)}dfrac{4xy^2}{x^2+3y^4} = lim_{(x,y)→(0,0)}dfrac{4(y^2)y^2}{(y^2)^2+3y^4} = lim_{(x,y)→(0,0)}dfrac{4y^4}{y^4+3y^4} = lim_{(x,y)→(0,0)}1 = 1. end{align*}]

By the same logic in part a, it is impossible to find a δ disk around the origin that satisfies the definition of the limit for any value of (ε<1.) Therefore,

[displaystyle lim_{(x,y)→(0,0)}dfrac{4xy^2}{x^2+3y^4} onumber]

does not exist.

Exercise (PageIndex{2}):

Show that

[lim_{(x,y)→(2,1)}dfrac{(x−2)(y−1)}{(x−2)^2+(y−1)^2} onumber]

does not exist.

Hint

Pick a line with slope (k) passing through point ((2,1).)

Answer

If (y=k(x−2)+1,) then (lim_{(x,y)→(2,1)}dfrac{(x−2)(y−1)}{(x−2)^2+(y−1)^2}=dfrac{k}{1+k^2}). Since the answer depends on (k,) the limit fails to exist.

Interior Points and Boundary Points

To study continuity and differentiability of a function of two or more variables, we first need to learn some new terminology.

Definition: interior and boundary points

Let (S) be a subset of (mathbb{R}^2) (Figure (PageIndex{4})).

  1. A point (P_0) is called an interior point of (S) if there is a (δ) disk centered around (P_0) contained completely in (S).
  2. A point (P_0) is called a boundary point of (S) if every (δ) disk centered around (P_0) contains points both inside and outside (S).

Definition: Open and closed sets

Let (S) be a subset of (mathbb{R}^2) (Figure (PageIndex{4})).

  1. (S) is called an open set if every point of (S) is an interior point.
  2. (S) is called a closed set if it contains all its boundary points.

An example of an open set is a (δ) disk. If we include the boundary of the disk, then it becomes a closed set. A set that contains some, but not all, of its boundary points is neither open nor closed. For example if we include half the boundary of a (δ) disk but not the other half, then the set is neither open nor closed.

Definition: connected sets and Regions

Let (S) be a subset of (mathbb{R}^2) (Figure (PageIndex{4})).

  1. An open set (S) is a connected set if it cannot be represented as the union of two or more disjoint, nonempty open subsets.
  2. A set (S) is a region if it is open, connected, and nonempty.

The definition of a limit of a function of two variables requires the (δ) disk to be contained inside the domain of the function. However, if we wish to find the limit of a function at a boundary point of the domain, the (δ) disk is not contained inside the domain. By definition, some of the points of the (δ) disk are inside the domain and some are outside. Therefore, we need only consider points that are inside both the (δ) disk and the domain of the function. This leads to the definition of the limit of a function at a boundary point.

Definition

Let (f) be a function of two variables, (x) and (y), and suppose ((a,b)) is on the boundary of the domain of (f). Then, the limit of (f(x,y)) as ((x,y)) approaches ((a,b)) is (L), written

[lim_{(x,y)→(a,b)}f(x,y)=L,]

if for any (ε>0,) there exists a number (δ>0) such that for any point ((x,y)) inside the domain of (f) and within a suitably small distance positive (δ) of ((a,b),) the value of (f(x,y)) is no more than (ε) away from (L) (Figure (PageIndex{2})). Using symbols, we can write: For any (ε>0), there exists a number (δ>0) such that

[|f(x,y)−L|<ε, ext{whenever}, 0

Example (PageIndex{3}): Limit of a Function at a Boundary Point

Prove

[lim_{(x,y)→(4,3)}sqrt{25−x^2−y^2}=0. onumber]

Solution

The domain of the function (f(x,y)=sqrt{25−x^2−y^2}) is (ig{(x,y)∈mathbb{R}^2∣x^2+y^2≤25ig}), which is a circle of radius (5) centered at the origin, along with its interior as shown in Figure (PageIndex{5}).

We can use the limit laws, which apply to limits at the boundary of domains as well as interior points:

[egin{align*} lim_{(x,y)→(4,3)}sqrt{25−x^2−y^2} =sqrt{lim_{(x,y)→(4,3)}(25−x^2−y^2)} = sqrt{lim_{(x,y)→(4,3)}25−lim_{(x,y)→(4,3)}x^2−lim_{(x,y)→(4,3)}y^2} =sqrt{25−4^2−3^2} = 0 end{align*}]

See the following graph.

Exercise (PageIndex{3})

Evaluate the following limit:

[lim_{(x,y)→(5,−2)}sqrt{29−x^2−y^2}. onumber]

Hint

Determine the domain of (f(x,y)=sqrt{29−x^2−y^2}).

Answer

[lim_{(x,y)→(5,−2)}sqrt{29−x^2−y^2} onumber]

Continuity of Functions of Two Variables

In Continuity, we defined the continuity of a function of one variable and saw how it relied on the limit of a function of one variable. In particular, three conditions are necessary for (f(x)) to be continuous at point (x=a)

  1. (f(a)) exists.
  2. (displaystyle lim_{x→a}f(x)) exists.
  3. (displaystyle lim_{x→a}f(x)=f(a).)

These three conditions are necessary for continuity of a function of two variables as well.

Definition: continuous Functions

A function (f(x,y)) is continuous at a point ((a,b)) in its domain if the following conditions are satisfied:

  1. (f(a,b)) exists.
  2. (displaystyle lim_{(x,y)→(a,b)}f(x,y)) exists.
  3. (displaystyle lim_{(x,y)→(a,b)}f(x,y)=f(a,b).)

Example (PageIndex{4}): Demonstrating Continuity for a Function of Two Variables

Show that the function

[f(x,y)=dfrac{3x+2y}{x+y+1} onumber]

is continuous at point ((5,−3).)

Solution

There are three conditions to be satisfied, per the definition of continuity. In this example, (a=5) and (b=−3.)

1. (f(a,b)) exists. This is true because the domain of the function f consists of those ordered pairs for which the denominator is nonzero (i.e., (x+y+1≠0)). Point ((5,−3)) satisfies this condition. Furthermore,

[f(a,b)=f(5,−3)=dfrac{3(5)+2(−3)}{5+(−3)+1}=dfrac{15−6}{2+1}=3. onumber]

2. (displaystyle lim_{(x,y)→(a,b)}f(x,y)) exists. This is also true:

[egin{align*} lim_{(x,y)→(a,b)}f(x,y) =lim_{(x,y)→(5,−3)}dfrac{3x+2y}{x+y+1} =dfrac{displaystyle lim_{(x,y)→(5,−3)}(3x+2y)}{displaystyle lim_{(x,y)→(5,−3)}(x+y+1)} = dfrac{15−6}{5−3+1} = 3. end{align*} onumber]

3. (displaystyle lim_{(x,y)→(a,b)}f(x,y)=f(a,b).) This is true because we have just shown that both sides of this equation equal three.

Exercise (PageIndex{4})

Show that the function

[f(x,y)=sqrt{26−2x^2−y^2} onumber ]

is continuous at point ((2,−3)).

Hint

Use the three-part definition of continuity.

Answer
  1. The domain of (f) contains the ordered pair ((2,−3)) because (f(a,b)=f(2,−3)=sqrt{16−2(2)^2−(−3)^2}=3)
  2. (displaystyle lim_{(x,y)→(a,b)}f(x,y)=3)
  3. (displaystyle lim_{(x,y)→(a,b)}f(x,y)=f(a,b)=3)

Continuity of a function of any number of variables can also be defined in terms of delta and epsilon. A function of two variables is continuous at a point ((x_0,y_0)) in its domain if for every (ε>0) there exists a (δ>0) such that, whenever (sqrt{(x−x_0)^2+(y−y_0)^2}<δ) it is true, (|f(x,y)−f(a,b)|<ε.) This definition can be combined with the formal definition (that is, the epsilon–delta definition) of continuity of a function of one variable to prove the following theorems:

The Sum of Continuous Functions Is Continuous

If (f(x,y)) is continuous at ((x_0,y_0)), and (g(x,y)) is continuous at ((x_0,y_0)), then (f(x,y)+g(x,y)) is continuous at ((x_0,y_0)).

The Product of Continuous Functions Is Continuous

If (g(x)) is continuous at (x_0) and (h(y)) is continuous at (y_0), then (f(x,y)=g(x)h(y)) is continuous at ((x_0,y_0).)

The Composition of Continuous Functions Is Continuous

Let (g) be a function of two variables from a domain (D⊆mathbb{R}^2) to a range (R⊆R.) Suppose (g) is continuous at some point ((x_0,y_0)∈D) and define (z_0=g(x_0,y_0)). Let f be a function that maps (R) to (R) such that (z_0) is in the domain of (f). Last, assume (f) is continuous at (z_0). Then (f∘g) is continuous at ((x_0,y_0)) as shown in Figure (PageIndex{7}).

Let’s now use the previous theorems to show continuity of functions in the following examples.

Example (PageIndex{5}): More Examples of Continuity of a Function of Two Variables

Show that the functions (f(x,y)=4x^3y^2) and (g(x,y)=cos(4x^3y^2)) are continuous everywhere.

Solution

The polynomials (g(x)=4x^3) and (h(y)=y^2) are continuous at every real number, and therefore by the product of continuous functions theorem, (f(x,y)=4x^3y^2) is continuous at every point ((x,y)) in the (xy)-plane. Since (f(x,y)=4x^3y^2) is continuous at every point ((x,y)) in the (xy)-plane and (g(x)=cos x) is continuous at every real number (x), the continuity of the composition of functions tells us that (g(x,y)=cos(4x^3y^2)) is continuous at every point ((x,y)) in the (xy)-plane.

Exercise (PageIndex{5})

Show that the functions (f(x,y)=2x^2y^3+3) and (g(x,y)=(2x^2y^3+3)^4) are continuous everywhere.

Hint

Use the continuity of the sum, product, and composition of two functions.

Answer

The polynomials (g(x)=2x^2) and (h(y)=y^3) are continuous at every real number; therefore, by the product of continuous functions theorem, (f(x,y)=2x^2y^3) is continuous at every point ((x,y)) in the (xy)-plane. Furthermore, any constant function is continuous everywhere, so (g(x,y)=3) is continuous at every point ((x,y)) in the (xy)-plane. Therefore, (f(x,y)=2x^2y^3+3) is continuous at every point ((x,y)) in the (xy)-plane. Last, (h(x)=x^4) is continuous at every real number (x), so by the continuity of composite functions theorem (g(x,y)=(2x^2y^3+3)^4) is continuous at every point ((x,y)) in the (xy)-plane.

Functions of Three or More Variables

The limit of a function of three or more variables occurs readily in applications. For example, suppose we have a function (f(x,y,z)) that gives the temperature at a physical location ((x,y,z)) in three dimensions. Or perhaps a function (g(x,y,z,t)) can indicate air pressure at a location ((x,y,z)) at time (t). How can we take a limit at a point in (mathbb{R}^3)? What does it mean to be continuous at a point in four dimensions?

The answers to these questions rely on extending the concept of a (δ) disk into more than two dimensions. Then, the ideas of the limit of a function of three or more variables and the continuity of a function of three or more variables are very similar to the definitions given earlier for a function of two variables.

Definition: (δ)-balls

Let ((x_0,y_0,z_0)) be a point in (mathbb{R}^3). Then, a (δ)-ball in three dimensions consists of all points in (mathbb{R}^3) lying at a distance of less than (δ) from ((x_0,y_0,z_0)) —that is,

[ig{(x,y,z)∈mathbb{R}^3∣sqrt{(x−x_0)^2+(y−y_0)^2+(z−z_0)^2}<δig}.]

To define a (δ)-ball in higher dimensions, add additional terms under the radical to correspond to each additional dimension. For example, given a point (P=(w_0,x_0,y_0,z_0)) in (mathbb{R}^4), a (δ) ball around (P) can be described by

[ig{(w,x,y,z)∈mathbb{R}^4∣sqrt{(w−w_0)^2+(x−x_0)^2+(y−y_0)^2+(z−z_0)^2}<δig}.]

To show that a limit of a function of three variables exists at a point ((x_0,y_0,z_0)), it suffices to show that for any point in a (δ) ball centered at ((x_0,y_0,z_0)), the value of the function at that point is arbitrarily close to a fixed value (the limit value). All the limit laws for functions of two variables hold for functions of more than two variables as well.

Example (PageIndex{6}): Finding the Limit of a Function of Three Variables

Find

[lim_{(x,y,z)→(4,1,−3)}dfrac{x^2y−3z}{2x+5y−z}. onumber]

Solution

Before we can apply the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, the identity law, and the constant law,

[egin{align*}lim_{(x,y,z)→(4,1,−3)}(2x+5y−z) =2(lim_{(x,y,z)→(4,1,−3)}x)+5(lim_{(x,y,z)→(4,1,−3)}y)−(lim_{(x,y,z)→(4,1,−3)}z) = 2(4)+5(1)−(−3) = 16. end{align*}]

Since this is nonzero, we next find the limit of the numerator. Using the product law, power law, difference law, constant multiple law, and identity law,

[egin{align*} lim_{(x,y,z)→(4,1,−3)}(x^2y−3z) =(lim_{(x,y,z)→(4,1,−3)}x)^2(lim_{(x,y,z)→(4,1,−3)}y)−3lim_{(x,y,z)→(4,1,−3)}z =(4^2)(1)−3(−3) = 16+9 = 25 end{align*}]

Last, applying the quotient law:

[lim_{(x,y,z)→(4,1,−3)}dfrac{x^2y−3z}{2x+5y−z}=dfrac{displaystyle lim_{(x,y,z)→(4,1,−3)}(x^2y−3z)}{displaystyle lim_{(x,y,z)→(4,1,−3)}(2x+5y−z)}=dfrac{25}{16} onumber]

Exercise (PageIndex{6})

Find

[lim_{(x,y,z)→(4,−1,3)}sqrt{13−x^2−2y^2+z^2} onumber]

Hint

Use the limit laws and the continuity of the composition of functions.

Answer

[lim_{(x,y,z)→(4,−1,3)}sqrt{13−x^2−2y^2+z^2}=2 onumber]

Key Concepts

  • To study limits and continuity for functions of two variables, we use a (δ) disk centered around a given point.
  • A function of several variables has a limit if for any point in a (δ) ball centered at a point (P), the value of the function at that point is arbitrarily close to a fixed value (the limit value).
  • The limit laws established for a function of one variable have natural extensions to functions of more than one variable.
  • A function of two variables is continuous at a point if the limit exists at that point, the function exists at that point, and the limit and function are equal at that point.

Glossary

boundary point
a point (P_0) of (R) is a boundary point if every (δ) disk centered around (P_0) contains points both inside and outside (R)
closed set
a set (S) that contains all its boundary points
connected set
an open set (S) that cannot be represented as the union of two or more disjoint, nonempty open subsets
(δ) disk
an open disk of radius (δ) centered at point ((a,b))
(δ) ball
all points in (mathbb{R}^3) lying at a distance of less than (δ) from ((x_0,y_0,z_0))
interior point
a point (P_0) of (mathbb{R}) is a boundary point if there is a (δ) disk centered around (P_0) contained completely in (mathbb{R})
open set
a set (S) that contains none of its boundary points
region
an open, connected, nonempty subset of (mathbb{R}^2)

14.2: Limits and Continuity

Thomas Calculus, 11th edition, pages: 965 - 995

Chapter 14 Partial Derivatives

The derivatives of functions of several variables are more varied and more interesting because of the different ways in which the variables can interact. Their integrals lead to a greater variety of applications. The studies of probability, statistics, fluid dynamics, and electricity, to mention only a few, all lead in natural ways to functions of more than one variable.

Sec 14.1 Function of Several Variables

In studying a real-world phenomenon, a quantity being investigated usually depends on two or more independent variables. So we need to extend the basic ideas of the calculus of functions of a single variable to functions of several variables. Although the calculus rules remain essentially the same, the calculus is even richer.

Sec 14.2 Limits and Continuity in Higher Dimension

This section treats limits and continuity for multivariable functions. The definition of the limit of a function of two or three variables is similar to the definition of the limit of a function of a single variable but with a crucial difference.

Sec 14.3 Partial Derivatives

The calculus of several variables is basically single-variable calculus applied to several variables one at a time. When we hold all but one of the independent variables of a function constant and differentiate with respect to that one variable, we get a &ldquopartial&rdquo derivative. This section shows how partial derivatives are defined and interpreted geometrically, and how to calculate them by applying the rules for differentiating functions of a single variable.


Solutions for Chapter 14.2: LIMITS AND CONTINUITY

Solutions for Chapter 14.2: LIMITS AND CONTINUITY

  • 14.2.1: Suppose that . What can you say about the value of ? What if is con.
  • 14.2.2: Explain why each function is continuous or discontinuous. (a) The o.
  • 14.2.3: Use a table of numerical values of for near the origin to make a co.
  • 14.2.4: Use a table of numerical values of for near the origin to make a co.
  • 14.2.5: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.6: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.7: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.8: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.9: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.10: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.11: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.12: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.13: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.14: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.15: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.16: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.17: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.18: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.19: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.20: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.21: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.22: Find the limit, if it exists, or show that the limit does not exist.
  • 14.2.23: Use a computer graph of the function to explain why the limit does .
  • 14.2.24: Use a computer graph of the function to explain why the limit does .
  • 14.2.25: Find and the set on which is continuous.
  • 14.2.26: Find and the set on which is continuous.
  • 14.2.27: Graph the function and observe where it is discontinuous. Then use .
  • 14.2.28: Graph the function and observe where it is discontinuous. Then use .
  • 14.2.29: Determine the set of points at which the function is continuous.
  • 14.2.30: Determine the set of points at which the function is continuous.
  • 14.2.31: Determine the set of points at which the function is continuous.
  • 14.2.32: Determine the set of points at which the function is continuous.
  • 14.2.33: Determine the set of points at which the function is continuous.
  • 14.2.34: Determine the set of points at which the function is continuous.
  • 14.2.35: Determine the set of points at which the function is continuous.
  • 14.2.36: Determine the set of points at which the function is continuous.
  • 14.2.37: Determine the set of points at which the function is continuous.
  • 14.2.38: Determine the set of points at which the function is continuous.
  • 14.2.39: Use polar coordinates to find the limit. [If are polar coordinates .
  • 14.2.40: Use polar coordinates to find the limit. [If are polar coordinates .
  • 14.2.41: Use polar coordinates to find the limit. [If are polar coordinates .
  • 14.2.42: At the beginning of this section we considered the function and gue.
  • 14.2.43: Graph and discuss the continuity of the function
  • 14.2.44: Let (a) Show that as along any path through of the form with . (b) .
  • 14.2.45: Show that the function given by is continuous on . [Hint: Consider .]
  • 14.2.46: If , show that the function f given by is continuous on .
Textbook: Multivariable Calculus,
Edition: 7
Author: James Stewart
ISBN: 9780538497879

Since 46 problems in chapter 14.2: LIMITS AND CONTINUITY have been answered, more than 38365 students have viewed full step-by-step solutions from this chapter. Chapter 14.2: LIMITS AND CONTINUITY includes 46 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. Multivariable Calculus, was written by and is associated to the ISBN: 9780538497879. This textbook survival guide was created for the textbook: Multivariable Calculus,, edition: 7.

Measure of the clockwise angle that the line of travel makes with due north

The central point in a circle, ellipse, hyperbola, or sphere

A branch of mathematics related to determining the number of elements of a set or the number of ways objects can be arranged or combined

Complex numbers a + bi and a - bi

The measure of an angle in degrees, minutes, and seconds

Reciprocal of the period of a sinusoid.

See Magnitude of a vector.

The perpendicular bisector of the major axis of an ellipse with endpoints on the ellipse.


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Teaching Calculus

My choices for the Good Question series are somewhat eclectic. Some are chosen because they are good, some because they are bad, some because I learned something from them, some because they can be extended, and some because they can illustrate some point of mathematics.

These posts contain a discussion of the questions and suggestions for using and adapting them. After writing about several questions, I hit upon the name “Good Question,” so while some have other names they are all interesting. The annotations give details and topics related to the question and the course description. They are in no particular order.

BEFORE CALCULUS

My Favorite Function (7-31-13) A pre-calculus question on finding roots and why calculators can’t do this one.

A Problem with 4 Solutions and 2 Morals. (6-6-2014). On simplifying an expression with radicals. Pre-calculus, calculator use.

LIMITS AND CONTINUITY

Good Question 5: 1998 AB 2 (7-29-2015) End behavior, limits at infinity, max-min, range, family of function. The correct use of infinity and DNE. Some students found a totally not expected way to do the last part without using calculus. A question I used in almost every presentation to teachers.

Dominance (8-8-2012) and Far Out (10-31-2012) A really fun limit to investigate after student know how to find extreme values , end behavior , and points of inflection

DERIVATIVES

At Just the Right Time (9-12-2013) A simple textbook question that I assigned before the class was ready for it with good results. slope, tangent line.

A Standard Problem? (5-14-2013) Find the closes point on a parabola. And then go from there. An investigation. Max-min . Also see The Marble and the Vase below for more on the same problem.

Mean Tables (9-16-2014) A discussion of 2003 AB 90 in which students which short table of values could be those for a function describes in the stem. Mean Value Theorem, Graph analysis.

The Marble and the Vase (10-23-2014) A further discussion of “A Standard Problem? ” – see above. Max-min, with graphs.

Related Rate Problems II (10-10-2012) Two problems you won’t find in textbooks

Good Question 1 2008 AB 6 (1-21-2015) Tangent lines, critical points, point of inflection and limit at infinity, relationship between f, f’ and f’’.

Good Question 3: 1995 BC 5 (7-8-2105) A questions designed for calculator solution for the first year graphing calculators were required on the exams. It a shame there were not more like this. Second derivative, concavity.

Good Question 7: 2009 AB 3 (10-5-2015) The “Mighty Cable Company” question, Accumulation (of money), how to determine meaning of a definite integral, max-min.

Determining the Indeterminate (12-6-2015) Implicit differentiation and analysis of a curve where the derivative is 0 and where it is an indeterminate form and what it all means.

Good Question 9 (2-14-2106) A related rate question that got me thinking. Max/min

Good Question 10: The Cone Problem (11-8-2016) Cutting a sector from a circle and making a cone of maximum volume. Domain of the model.

Good Question 11 – Riemann Reversed (11-29-2016) Given a Riemann sum find the associated function and its domain so you can find the integral. This is the reverse of the usual problem when one finds the Riemann sum first and is becoming a common question on the AP Calculus Exams. There are several examples and a discussion of the concern that the answer is never unique, which makes it a poor question.

Good Question 4: 2008 AB 10 (7-15-2015) A multiple-choice question on comparing several Riemann sums: left-, right-, midpoint, and trapezoid sums. Lots of ways to extend and adapt for your class.

Most Triangles are Obtuse! (1-18-2013) Using the average value of a function concept a 10 th grade BC student proved this fact. Integration .

Challenge and Solution (3-29-2013 and 4-8-2013) Find when a vase is half-full. Two different methods with two very different correct solution. Why? Volume, washer method, shell method .

Variations on a Theme by ETS (6-14-2013) Adapting simple multiple-choice question (2008 AB 9) on accumulation and definite integration

Variations on a Theme – 2 (6-28-2013) Riemann sums and ideas on how to adapt this problem to get more out pf int. Also, discussed under Good Question 4: 2008 AB 10 below.

Lin McMullin’s Theorem (7-10-2013) A sighting of the Golden Ratio in the points of inflection of any quartic polynomial and More Gold (7-17-2013) another sighting of the Gold Ratio in cubic polynomials by David Tschappat

Half-full and Half-empty (1-16-2015) A quick thought experiment leading to the concept of the average value of a function.

Good Question 6: 2000 – AB 4 (8-25-2015) Another of my favorite questions for working with teachers and teaching accumulation. Finding the function from its derivative done two ways.

Good Question 8 or not? (1-5-2016) A textbook question that is not good, but from which you can learn a lot. Unit analysis, accumulation, CAS work, approximation (and not a good one), reading a Riemann sum and its units, periodic functions.

Good Question 12 – Parts with a Constant (12-13-2016) You’re always telling students, “Don’t forget the + C, the constant of integration.” So now we do integration by parts and don’t worry about the + C. Why?

Good Question 13 (12-12-2017) An antiderivative 4 ways: u-substitution, integration by parts, a different u-substitution and adding zero in a convenient form. All are correct, but the answers are not the same, or are they?

Good Question 15: 2018 BC 2(a) (5-23-2018) A accumulation question with rather strange units. Making sense of the units. A BC question suitable for AB.

DIFFERENTIAL EQUATIONS

Good Question 2: 2002 BC 2 (2-17-2015) Differential equations: Suitable for AB: slope field, particular solutions, max-min, second derivative test, a clever solution, and for BC only: Euler’s Method. Continued in A Family of Functions next below.

A Family of Functions (2-21-2015) Jumping off from Good Question 2: 2002 BC 2 immediately above, a look at all the solutions. End behavior, maximum points, graphing.

Euler’s Method for Making Money (2-25-2015) Relating exponential growth and compound interest to Euler’s Method.

Good Question 16: 2018 BC 2(b) (5-29-2018) A density problem. Using units to find the solution. A BC question suitable for AB.

Summer Fun (6-12-18) links to an exploration on a question similar to 2018 AB 6. Finding the general solution of the differential equation by separating the variables, c hecking the solution by substitution, u sing a graphing utility to explore the solutions for all values of the constant of integration, C, finding the solutions’ horizontal and vertical asymptotes, finding several particular solutions, finding the domains of the particular solutions, f inding the extreme value of all solutions in terms of C, f inding the second derivative (implicit differentiation), c onsidering concavity, i nvestigating a special case or two. The exploration is here and the solutions are here.

SEQUENCES AND SERIES

Good Question 14 (2-23-2018) The Integral Test for convergence or divergence of an infinite series. Why it works. (Similar to 2016 BC 92)


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SAP BO/BI CMS Server Installation

  • Login to CMS server (azwinbocms1) as administrator.
  • Setup ODBC connection with Azure SQL DB
    • Download ODBC driver version 13.1 and install on CMS server.

      • Goto to Control Panel -> Administrative Tools -> ODBC Datasource (64-bit) -> System DSN. Click on ‘Add’ to create connection for CMS database.

      Enter the name of the Azure SQL server

      Enter the server admin user and password.

      Select the database name from the drop-down list.

      Test the connection and click on ‘Finish’.

        • Repeat the above steps to create connection for AUDIT database.
        • Install CMS Server
          • Run setup.exe.
          • Select the Language Package. Default is English.
          • Select Custom/Expand Option to choose the application to be installed.
          • Choose the location of the installation. Installing it on I drive.
          • In the ‘Select Feature’ screen, Uncheck the ‘Web Tier’ and ‘Sybase SQL Anywhere DB’.
          • Choose the Option, Start a New SAP BO/BI platform deployment.
          • Select DB type for CMS Repository: MS SQL Server using ODBC.
          • Select DB type for Audit DB: MS SQL Server using ODBC.
          • Continue with Installation/configuration parametes.
          • In the screen to enter Configuration details for CMS & Audit DBs. Check the DB server & DB name. Enter the DB username & password.
          • In next screen, continue with entering other configuration details and CMS Installation.

          14.10 Track Warrant in Effect

          A track warrant is in effect until a crew member reports the train has cleared the limits, or the track warrant is made void. The crew member must inform the train dispatcher when the train has cleared the limits.

          Time Limit Shown

          If the track warrant shows a time limit, the train must clear the limits by the time specified, unless another track warrant is obtained. If the crew members cannot contact the train dispatcher and time limits expire, authority is extended until the train dispatcher can be contacted.


          Definition of 'Graph Theory'

          Definition: Graph is a mathematical representation of a network and it describes the relationship between lines and points. A graph consists of some points and lines between them. The length of the lines and position of the points do not matter. Each object in a graph is called a node.

          Description: A graph ‘G’ is a set of vertex, called nodes ‘v’ which are connected by edges, called links ‘e'. Thus G= (v , e).
          Vertex (Node): A node v is an intersection point of a graph. It denotes a location such as a city, a road intersection, or a transport terminal (stations, harbours, and airports).

          Edge (Link): An edge e is a link between two nodes. A link denotes movements between nodes. It has a direction that is generally represented as an arrow. If an arrow is not used, it means the link is bi-directional.

          Transport geography can be defined by a graph. Most networks, namely road, transit, and rail networks, are defined more by their links than by nodes. But it is not true for all transportation networks. For instance, air networks are defined more by their nodes than by their links since links are mostly not clearly defined. A telecommunication system can also be represented as a network. Mobile telephone networks or the internet is the considered the most complex graph. However, cell phones and antennas can be represented as nodes whereas links could be individual phone calls. The core of the internet or servers can also be represented as nodes while the physical infrastructure between them, like fiber optic cables, can act as links. This suggests that all transport networks can be represented by graph theory in some way.


          Patient Counseling Information

          Precedex is indicated for short-term intravenous sedation. Dosage must be individualized and titrated to the desired clinical effect. Blood pressure, heart rate and oxygen levels will be monitored both continuously during the infusion of Precedex and as clinically appropriate after discontinuation.

          • When Precedex is infused for more than 6 hours, patients should be informed to report nervousness, agitation, and headaches that may occur for up to 48 hours.
          • Additionally, patients should be informed to report symptoms that may occur within 48 hours after the administration of Precedex such as: weakness, confusion, excessive sweating, weight loss, abdominal pain, salt cravings, diarrhea, constipation, dizziness or light-headedness.

          Distributed by Hospira, Inc. Lake Forest, IL 60045


          Watch the video: Όριο - Συνέχεια: Παράδειγμα 3 (December 2021).