Solving Limits with Delta-Epsilon Proofs

[ineedhelpnow]

PLEASE HELP! i am so lost on this. we're using delta epsilon proofs and i am so confused since it was never properly taught to me in calc 1.

find the limit.
$\lim_{{(x,y)}\to{(0,0)}}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$

 

[MarkFL]

I'm thinkin' you should convert to polar coordinates...

Usually you are given the value of the limit in an epsilon-delta proof. To simply find the value of this limit, convert to polars, then use L'Hôpital. What do you find?

 

[ineedhelpnow]

im confused on how to find the limit of this...

 

[MarkFL]

im confused on how to find the limit of this...

Converting to polar coordinates means to use:

$$r^2=x^2+y^2$$

Then you will have a limit in one variable...

 

[ineedhelpnow]

im lookin at the way my book would do it and first that set f(x,y)=(whatever your taking the limit of) and then they do f(x,0) (which I am stuck on) and f(0,y). by using polar coordinates, wouldn't i end up with a 0 in the denominator?

 

[MarkFL]

im lookin at the way my book would do it and first that set f(x,y)=(whatever your taking the limit of) and then they do f(x,0) (which I am stuck on) and f(0,y). by using polar coordinates, wouldn't i end up with a 0 in the denominator?

You will end up with the indeterminate form 0/0, and so this is why I suggested our friend L'Hôpital. :D

 

[ineedhelpnow]

oh i got it using another method. the answer is 2. is that the final answer though?

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i think the way my book wants me to do it is by testing it with several functions to see if it all approaches the same limit.

 

[MarkFL]

Yes, 2 is what I got...

$$\lim_{r\to0}\frac{r^2}{\sqrt{r^2+1}-1}=\lim_{r\to0}\frac{2r}{\dfrac{2r}{2\sqrt{r^2+1}}}=\lim_{r\to0}2\sqrt{r^2+1}=2$$

edit: I wanted to add that since when we converted to polar coordinates, and there was no $\theta$, we know we are approaching the limit from "all angles." :D