# Lim f(0,y). y->0. L'H rule needed, did i do it right?

1. Oct 11, 2005

### mr_coffee

Hello everyone. I think i did this right but i want to make sure. Let f(x,y) = (x^2+sin^2(y))/(2x^2+y^2); I want to find the limit as y ->0;
Find lim y->0 f(0,y).
Here is my work:
http://img204.imageshack.us/img204/194/lastscan4zi.jpg [Broken]

Also the 2nd part to this problem is the following, and here is my work:
http://show.imagehosting.us/show/790806/0/nouser_790/T0_-1_790806.jpg
Do you think that is correct? thanks.

Last edited by a moderator: May 2, 2017
2. Oct 11, 2005

### iNCREDiBLE

You seem to have misunderstood L'Hospitals rule.

L'Hospitals rule states that $\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}$ if $\lim f(x)$ and $\lim g(x)$ are both zero or ±$\infty$. L'Hospitals rule isn't necessary. Just use the standard limit for $\frac{sin(x)}{x}$

3. Oct 11, 2005

### mr_coffee

I havn't had calc I in awhile, my professor said L'Hopsitals rule is needed...hm...
so the sin^2(y)/y^2 is just really 1 then?

4. Oct 12, 2005

### Pseudo Statistic

For the first one you differentiated incorrectly... use the chain rule or product rule for the numerator.. (Whatever you feel more comfortable with)
The answer is indeed 1, yes.
No clue about the second one... I'm no epsilon-delta guy. :D

5. Oct 12, 2005

### HallsofIvy

Staff Emeritus
You could use L'Hopital's rule but you don't need to.
$$f(0,y)= \frac{sin^2(y)}{y^2}= \frac{sin(y)}{y}\frac{sin(y)}{y}$$ and you should know the limit of those!
As for your second question, your answer to part (c) simply asserts what you want to show with no proof. I take it that there was a part (a) that asked you to show that $$lim_{x->0}f(x,0)= \frac{1}{2}$$. Can you show that given some $$\epsilon$$> 0 and any $$\delta$$> 0 there a point (x,0) closer to (0,0) $$\delta$$ such that f(x,0) is closer to 1/2 than $$\epsilon$$- and another point (0,y) closer to (0,0) than $$\delta$$ such that f(0,y) is closer than 1 than $$\epsilon$$ and so there no single number that both are close to?

Last edited: Oct 12, 2005
6. Oct 12, 2005

### mr_coffee

This is what i ended up getting...
http://img410.imageshack.us/img410/6888/lastscan7rp.jpg [Broken]
look any better?

Last edited by a moderator: May 2, 2017