How to find limit of multivariable function?

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SUMMARY

The discussion focuses on finding the limit of multivariable functions, specifically analyzing two examples. The first limit, \(\lim_{(x,y)\to(0,0)}\frac{(xy\cos y)}{3x^2+y^2}\), is established to not exist due to differing limits along various paths. In contrast, the second limit, \(\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}\), is shown to exist and equal 0 using the Squeeze Theorem, despite initial confusion regarding the method of approach. The conversation emphasizes the importance of testing multiple paths and applying the Squeeze Theorem for verification.

PREREQUISITES
  • Understanding of multivariable limits
  • Familiarity with the Squeeze Theorem
  • Knowledge of evaluating limits along different paths
  • Basic calculus concepts related to continuity and limits
NEXT STEPS
  • Study the Squeeze Theorem in detail and its applications in multivariable calculus
  • Practice evaluating limits of multivariable functions using various paths
  • Explore techniques for proving the existence of limits in multivariable calculus
  • Review examples of limits that do not exist and the reasoning behind them
USEFUL FOR

Students studying multivariable calculus, educators teaching limit concepts, and anyone seeking to deepen their understanding of limit evaluation techniques in higher mathematics.

Puchinita5
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Homework Statement



Hi. I'm going to provide two homework examples. The first one I got correctly. But the second one, I used the same method as the first question but I got the answer incorrect. I'm really confused.

The first question was
[tex] \lim_{(x,y)\to(0,0)}\frac{(xycosy)}{3x^2+y^2}[/tex]
f(x,0)=0
f(0,y)=0
for y=x , f(x,x)= (1/4)cosx

since the three limits are not equal, the limit does not exist!

Okay now the second question:

[tex] \lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}[/tex]

I again did:
f(x,0)=0
f(0,y)=0
[tex]f(x,x)=\frac{x^2}{x\sqrt{2}}[/tex]

since they aren't the same, limit does not exist. I THOUGHT.
I looked in the solutions manual, and it says "We can see that the limit along any line through (0,0) is 0, as well as along other paths through (0,0) such as x=y^2. So we suspect that the limit exists and equals 0; we use the Squeeze Theorem to prove our assertion.

[tex] 0\leq\left | (\frac{xy}{\sqrt{x^2+y^2}})\right |\leq \left |x \right | since \left |y\right |\leq \sqrt(x^2+y^2) and \left |x\right | \to \0 \ as (x,y) \to \(0,0).\ so\ the\ limit\ equals\ 0.[/tex]

I guess I don't understand how you choose the method to solve for the limit. Is there something about the function that tells you what to do? Why is f(x,x) not an option?

I mean, I guess it makes sense because I'm sure you can't just plug in anything or at some point you will get something that won't equal the same thing. But how do you know what to do? How would I have known to use the squeeze theorem?

I'm so confused.
[
 
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this always happens to me...the SECOND i post something on this site i realize something.

I am supposed to plug in a zero into the (x^2)/(x2^.5) aren't I?
which would also make it equal to 0.

In which case...can someone just explain to me the logic behind the squeeze theorem in this example? It still confuses me.

And if you test out different functions, such as f(x,x) or f(x,0) etc... and get the same limit for all of them, do you always need to use the squeeze theorem to verify that it is indeed the limit?
 
Limits of functions of two variables can be very tricky. The problem is that there is not a standard approach that you can count on work any problem. So, given a problem where you don't know whether the limit exists, the usual thing is to check to see if you can get different answers along different paths. If you can, that is usually pretty easy and it settles the problem giving that the limit doesn't exist.

If you keep getting the same answer along many different paths, that may be because the limit does exist and is equal to that common number. Proving it is then where the difficulty usually arises. If you suspect, for example, that the limit is 0, then you can try to overestimate the function with something easier that does go to 0. That is the idea of the squeeze principle. You will learn a few techniques along the way to try on such problems after which they won't seem quite so intimidating. Don't get too hung up on it if you are having trouble with these at first. Experience and seeing more examples will help. And your course will soon move on to other more enjoyable topics.
 

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