I Multivariable limits

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1. Jul 29, 2016

mr.tea

I have found that multivariable limits are harder to find and/or prove that something exists.
Do you have any recommendations, given questions like "find(if exists) the limit....".

For example, I have no idea how to even start thinking about the following limit(if it exists or not, and if it does, what is the value):
$$\lim_{\textbf{x}\rightarrow 0} \frac{e^{|\textbf{x}|^2}-1}{|\textbf{x}|^2 +x_1^2x_2+x_2^2x_3}$$

where $$\textbf{x}=(x_1,x_2,x_3)$$

I would like to get any help for dealing with limits such as the aforementioned and others.

Thank you.

Last edited: Jul 29, 2016
2. Jul 29, 2016

micromass

Staff Emeritus
First things I would do:
1) Plot the function with a graph tool. This way you might see if the limit exists or not.
2) If the limit exists, then if $\mathbf{x}_n\rightarrow \mathbf{x}$, then $f(\mathbf{x}_n)\rightarrow \text{limit}$. So choose some simple sequences and see if they all yield the same limiting value.

Also, I'm not sure what $\mathbf{x}\rightarrow \infty$ means, can you define it?

3. Jul 29, 2016

BvU

Hi,

not a very good example: the $e^{\vec x^2}$ trumps everything. No limit.

In general:
focus on the heavies: the $-1$ in the numerator can be ignored.
if sensible, divide numerator and denominator by the heaviest term and see what's left over.
If what's left tends to 0/0 see which goes to 0 the fastest:
use taylor series
use l'Hopital's rule
google 'limit theorems' to find more !

4. Jul 29, 2016

Staff: Mentor

It could still go to infinity, which is not a limit in the real numbers but still some sort of limit. It does not, but that needs further analysis here.

5. Jul 29, 2016

micromass

Staff Emeritus
Am I the only one who doesn't know what $\mathbb{x}\rightarrow \infty$ means? How are you guys interpreting this?

6. Jul 29, 2016

mr.tea

Oops, it should go to 0. Sorry.
What would you do if you get this question on an exam or in a situation when you can't use a graphical tool? how would you proceed?
Probably the reason why the example isn't good is because $\mathbf{x}\rightarrow \infty$ should be $\mathbf{x}\rightarrow 0$.
How do you use l'Hopital's rule in multivariable calculus?

7. Jul 29, 2016

Staff: Mentor

If $\lim_{a \to \infty} \sup_{|x|=a} f(x) = \lim_{a \to \infty} \inf_{|x|=a} f(x)$ (assuming those things are well-defined), it makes sense to call this value "limit" I think.

For x -> 0 I would multiply numerator and denominator with the same thing to make the denominator nicely convergent (with a finite limit), which allows to simplify the expression.

8. Jul 29, 2016

BvU

Back to the example that was slightly modified to become more interesting. I would interpret $\vec x\downarrow \vec 0$ as: going to $(0,0,0)$ from any possible direction. (Must admit I never had any of these vector limits on an exam -- at least I don't remember ).

Taylor series is good for the numerator: $\exp (x_1^2 +x_2^2+x_3^2) - 1 = (x_1^2 +x_2^2+x_3^2) + \mathcal O\left ( (x_1^2 +x_2^2+x_3^2)^2 \; \right )$.
Divide numerator and denominator by $(x_1^2 +x_2^2+x_3^2)$ and get (1 + small stuff) / (1 + small stuff) $\Rightarrow$ limit exists and is 1.

9. Jul 29, 2016

micromass

Staff Emeritus
It makes sense, and this is the likely interpretation if you get something like this. But it is not the only possible interpretation. With $\mathbf{x}\rightarrow 0$ it is very unambigious. With "$\mathbf{x}\rightarrow \infty$, I think it should be defined first. But anyway, the problem has an error so it's not important.

Last edited by a moderator: Jul 29, 2016
10. Jul 29, 2016

mr.tea

Thank you. This is really becomes easy with Taylor series. I think I should use it more often in these situations.

I will collect your advice and try to do some more exercises. Still, if you have more suggestions, I would be grateful to hear them.

Thank you all.

11. Jul 29, 2016

Staff: Mentor

"From any possible direction" works only if you include curved "directions". There are functions where all straight lines lead to 1-dimensional functions that converge nicely, but the actual two-dimensional function does not converge. One example:
$$f(x) = \frac{xy}{x^2+y}$$

As bonus, it makes the three-dimensional limit one-dimensional, as we have to care about |x| only.

12. Jul 29, 2016

Staff: Mentor

I didn't know either. At latest when it comes to differentials such a sloppy notation will heavily strike back.

13. Jul 29, 2016

MAGNIBORO

I take this opportunity to make a basic question.
there is some method to determine if the limit exists?, and not try all possible ways to reach the indeterminate point.
example:

the limit is 0 but the method is try the path of x=0, y=0, y=mx , y=mx^2+nx.
there any way to know if the limit exists or not exist that does not involve check all the paths like a computer ?

14. Jul 30, 2016

mr.tea

My first thought when I saw this limit was moving to polar coordinates, because it has terms that look like $x^2+y^2$. So I just added in the denominator $+4x^2-4x^2(=0)$ to get something with the $5y^2$. Then I arrived to $r\cdot$(some finite number) which goes to 0 as $r$ goes to 0.

15. Jul 30, 2016

Staff: Mentor

There is an infinite set of paths, you cannot check them all.

Going to radial coordinates is often a good approach.

16. Jul 30, 2016

thanks =D