# 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

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

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

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