# Multi variable Limit

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1. Jul 12, 2018

### Scrope

1. The problem statement, all variables and given/known data
https://gyazo.com/268bef206850bfbf30fb0cca3f783599 <----- The question
2. Relevant equations

3. The attempt at a solution
Had this on a test today, honestly not sure how to evaluate. I know you can pass the limit to the inside of arctan but I can't see how the inside goes to infinity. Any help would be appreciated I just want to understand how to do it!

2. Jul 12, 2018

### scottdave

I don't get infinity for the inside, either. What is the answer that they gave? Did they get pi/2 for the limit? What answer did you get for the inside (and the limit)?

3. Jul 12, 2018

### Scrope

the answer given is pi/2 for the overall answer and +infinity for the inside

4. Jul 12, 2018

### Scrope

The only advice I have received so far is to take note that the bottom is the square of the distance from (0,1) but I can’t figure out what they mean

5. Jul 12, 2018

### Staff: Mentor

The numerator inside the brackets is approaching 1, and the denominator is approaching 0 from the positive side. It seems pretty clear to me that the part in brackets is approaching $+\infty$, so the overall limit is what?

6. Jul 13, 2018

### Scrope

the way I had learned it was that the only way to prove a limit existed was through methods such as delta-epsilon definition, switching to R or Rho, or by squeeze theorem. Is there any mathematical way to show that this goes to +infinity? We were told we were unable to evaluate limits by observation.

7. Jul 13, 2018

### Staff: Mentor

For the record, here is the limit in the image link you posted:
$$\text{Find }\lim_{(x, y) \to (0, 1)} \tan^{-1}\left[\frac{x^2 + 1}{x^2 + (y - 1)^2}\right]$$
The problem asks you to find the limit, but doesn't ask you to prove that this value is the limit. Proving the limit would require the use of a delta-epsilon proof or a theorem such as the squeeze theorem.

A major difficulty with these kinds of limits is when the limit is an indeterminate form such as $[\frac 0 0]$. This limit is not one of the indeterminate forms, since the numerator approaches 1 and the denominator approaches 0 through the positive numbers.

8. Jul 13, 2018

### Ray Vickson

Yes, the way to do it is to use the definition of the notion $(x,y) \to (0,1)$. Just using that definition (and a few simple manipulations), it is easy to show that for any (large) number $N > 0$, we know that as $(x,y)$ nears $(0,1)$ (in a well-defined sense that you are supposed to know) we have that the denominator of the inside fraction becomes $< 1/N$. The numerator is $\geq 1$, so the inside fraction is $> N$. That is essentially the definition of $\text{fraction} \to +\infty.$

9. Jul 13, 2018

### scottdave

Oh, that's easier to read than the tiny image. I misread what x and y we're approaching.

10. Jul 13, 2018

### epenguin

It doesn't matter whether the 'inside' is +∞ or -∞ , its tan-1 is π/2 in either case.

You appear to be able to approach the limit in several ways including from y>1.

The denominator is indeed the squared distance of a point (x, y) from (0,1).

11. Jul 14, 2018

### Staff: Mentor

Not so. $\lim_{x \to -\infty}\tan^{-1}(x) = -\frac \pi 2$. For this inverse trig function, the restricted-domain function $y = \tan(x), -\frac \pi 2 < x < \frac \pi 2$ is used.