# Evaluating limit for this function

• toforfiltum
In summary, the function ##f(x,y,z)= e^{xz} \cos (y^2) - x## approaches -1 as ##(x,y,z)## approaches ##(0,\sqrt{\pi},1)## from different paths, demonstrating the continuity of the function at that point.
toforfiltum

## Homework Statement

Function is ##lim_{(x,y,z) \rightarrow (0,\sqrt\pi,1)} \ e^{xz} \cos y^2 - x##

## The Attempt at a Solution

As ##x \rightarrow 0## along ##y= \sqrt \pi, z=1##, ##f(x,y,z)= -1##

As ##y \rightarrow 0## along ##x=0, z=1##, ##f(x,y,z) = -1##

As ##z \rightarrow 1## along ##x=0, y= \sqrt \pi##, ##f(x,y,z) = -1##

From what I have done, the limit looks like it is -1. I'm wondering if there are any other functions I could use to show that it is true. Or this is not needed at all, because of the argument that the function is continuous at the point ##(0,\sqrt \pi, 1)##? If so, the above steps are not needed at all, right?

Thanks.

Last edited:
toforfiltum said:

## Homework Statement

Function is ##lim_{(x,y,z) \rightarrow (0,0,0)} \ e^{xz} \cos y^2 - x##

## The Attempt at a Solution

As ##x \rightarrow 0## along ##y= \sqrt \pi, z=1##, ##f(x,y)= -1##

As ##y \rightarrow 0## along ##x=0, z=1##, ##f(x,y) = -1##

As ##z \rightarrow 1## along ##x=0, y= \sqrt \pi##, ##f(x,y) = -1##

From what I have done, the limit looks like it is -1. I'm wondering if there are any other functions I could use to show that it is true. Or this is not needed at all, because of the argument that the function is continuous at the point ##(0,\sqrt \pi, 1)##? If so, the above steps are not needed at all, right?
The function in the limit is continuous at (0, 0, 0), so there's not need to be concerned about different paths to this point. Also, why are you concerned about the point ##(0, \sqrt{\pi}, 1)##? It's nowhere near to the point at which the limit is being taken.

Mark44 said:
The function in the limit is continuous at (0, 0, 0), so there's not need to be concerned about different paths to this point. Also, why are you concerned about the point ##(0, \sqrt{\pi}, 1)##? It's nowhere near to the point at which the limit is being taken.
Oops, sorry! My typo. The question did indeed ask me to evaluate the limit at ##(0, \sqrt\pi, 1)##.

Thanks!

toforfiltum said:
Oops, sorry! My typo. The question did indeed ask me to evaluate the limit at ##(0, \sqrt\pi, 1)##.
OK, then that makes more sense, but it doesn't change what I said. The exponential part (##e^{xz}##) is continuous for every real value of x and z; the cosine function is continuous for all real y (and y2) and the simple polynomial x is continuous for all real x. Sums, differences, and products of these functions are also continuous for all real x, y, and z.

Where you run into problems is with quotients or other operations that produce undefined results.

Mark44 said:
OK, then that makes more sense, but it doesn't change what I said. The exponential part (##e^{xz}##) is continuous for every real value of x and z; the cosine function is continuous for all real y (and y2) and the simple polynomial x is continuous for all real x. Sums and differences of these functions are also continuous for all real x, y, and z.
So in this case, the limit at the point is equal to the inputs of the function.

Thanks!

toforfiltum said:

## Homework Statement

Function is ##lim_{(x,y,z) \rightarrow (0,\sqrt\pi,1)} \ e^{xz} \cos y^2 - x##

## The Attempt at a Solution

As ##x \rightarrow 0## along ##y= \sqrt \pi, z=1##, ##f(x,y)= -1##

As ##y \rightarrow 0## along ##x=0, z=1##, ##f(x,y) = -1##

As ##z \rightarrow 1## along ##x=0, y= \sqrt \pi##, ##f(x,y) = -1##

From what I have done, the limit looks like it is -1. I'm wondering if there are any other functions I could use to show that it is true. Or this is not needed at all, because of the argument that the function is continuous at the point ##(0,\sqrt \pi, 1)##? If so, the above steps are not needed at all, right?

Thanks.
It looks like there are some typos in your post.

From what's in the attempt, it appears that you actually are interested in ##\displaystyle\ lim _{(x,\,y,\,z)\to (0,\, \sqrt \pi ,\, 1)}\ f(x,y,z)\,,\ ## where ##\displaystyle\ f(x,y,z)= e^{xz} \cos (y^2) - x \ .##

Where you have ##\ f(x,y)\,,\ ## I think you mean to have ##\ f(x,y,z)\ .##(It took me wayyyyyy tooooo long to type this in !)

SammyS said:
It looks like there are some typos in your post.

From what's in the attempt, it appears that you actually are interested in ##\displaystyle\ lim _{(x,\,y,\,z)\to (0,\, \sqrt \pi ,\, 1)}\ f(x,y,z)\,,\ ## where ##\displaystyle\ f(x,y,z)= e^{xz} \cos (y^2) - x \ .##

Where you have ##\ f(x,y)\,,\ ## I think you mean to have ##\ f(x,y,z)\ .##(It took me wayyyyyy tooooo long to type this in !)
Oh gosh, yes. I meant ##f(x,y,z)##. So sorry for the carelessness. I guess my brain is too tired of multivariable calculus.

## 1. What is the purpose of evaluating limits for a function?

Evaluating limits for a function is important because it helps us understand the behavior of the function near a specific point. It also allows us to determine if the function is continuous at that point, and if not, how it approaches that point.

## 2. How do you evaluate a limit for a function algebraically?

To evaluate a limit for a function algebraically, you can start by substituting the value of the point into the function. If the resulting expression is undefined, then the limit does not exist. If the expression is defined, you can simplify it and see if it approaches a specific value as the point gets closer and closer to the given value.

## 3. Can a limit exist even if the function is not defined at that point?

Yes, a limit can exist even if the function is not defined at that point. This is known as a removable discontinuity, where the function has a hole at that point. In this case, the limit would be equal to the value of the function after the hole has been filled in.

## 4. What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the behavior of the function as the point approaches from one direction (either the left or the right). A two-sided limit, on the other hand, considers the behavior of the function as the point approaches from both directions. It is important to specify which type of limit is being evaluated, as they can have different values.

## 5. Can limits help us determine the continuity of a function?

Yes, limits can help us determine the continuity of a function. If the limit exists at a point and is equal to the function value at that point, then the function is continuous at that point. However, if the limit does not exist or is not equal to the function value, then the function is not continuous at that point.

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