# Having trouble applying squeeze theorem

1. Jul 29, 2014

### canon23

1. The problem statement, all variables and given/known data
Find the limit, if it exists, or show that the limit does
not exist:

lim (x,y) -> (0,0) [(y^2*sin(x)^2)/(x^4+y^4)]

(According to the textbook the limit 'does not exist')

3. The attempt at a solution

Since the function is approaching the origin [(0,0)]:

test path along y-axis:
let x = 0
lim (y) -> (0) [0/y^4] = 0

test path along x-axis:
let y = 0
lim (x) -> (0) [0/x^4] = 0

This is not efficient enough to prove that the limit is 0, therefore investigate further possibly using the squeeze theorem.

I know that [(y^2)/(x^4+y^4)] <= 1 (but i don't know what to do with this information)

0 <= [(y^2*sin(x)^2)/(x^4+y^4)] <= **

** Here is where i got stuck. I do not know what function fits the criteria and how to look for it. Please excuse me I just learned this theorem recently and I'm hoping someone can help me.

2. Jul 29, 2014

So you now know that the limit is zero if you approach $(0,0)$ from the x-axis (or y-axis). Can you find another path that will give you a nonzero limit?

3. Jul 29, 2014

### canon23

So far I've tried:

let y = x
let y = x^2
let y = x^3
let y = sqrt(x)

and got 0 for all. Is it possible to solve using the squeeze theorem?

4. Jul 29, 2014

Try $y=x$ again and show your work.

5. Jul 29, 2014

### canon23

let y = x

lim (x,x) -> (0,0) [(x^2*sin(x)^2)/(x^4+x^4)]
= lim (x) -> (0) [(x^2*sin(x)^2)/(2*x^4)]
= lim (x) -> (0) [(x^2*sin(x)^2)/(2*x^2*x^2)]
= lim (x) -> (0) [(sin(x)^2)/(2*x^2)]
= 0

6. Jul 29, 2014

Have another look at your last equality. Are you sure that
\begin{equation*}
\lim_{x \rightarrow 0} \frac{ \sin^2 x}{2x^2} = 0?
\end{equation*}

7. Jul 29, 2014

### canon23

Yes you're right I would get 0/0 (undefined) and the limit does not exist. The other paths I mentioned early turn out to be the same. Thanks for your help. Could this be solved using the squeeze theorem?

8. Jul 30, 2014

No. The limit
\begin{equation*}
\lim_{x \rightarrow 0} \frac{\sin^2 x}{2x^2}
\end{equation*}
DOES exist. It is just not zero! You should compute it. You could use the fact that you know (?) $\lim_{x \rightarrow 0} \frac{\sin x}{x}$ or use something powerful like a Taylor expansion or l'Hopital.

9. Jul 30, 2014

### vela

Staff Emeritus
Not really. The squeeze theorem allows you to prove that a function has a limit, but in this case, you're trying to show that the limit of a given function does not exist.