Limit involving delta-epsilon proof

[ZPlayer]

Hi, Everyone,

Problem asks to prove that limit of x^2 * sin^2 (y) / (x^2 + 2* y^2) as (x,y) approach (0,0) is 0 using delta-epsilon method. Please let me know how to solve, very complicated.

Thanks.

 

[gabbagabbahey]

What have you tried so far?

 

[ZPlayer]

What I tried so far is here.

 

[gabbagabbahey]

What I tried so far is here.

The condition that you need to satisfy is that

$$|f(x,y)-0|=\left| \frac{x^2sin^2(y)}{x^2+2y^2} \right| \leq \varepsilon$$

So showing that $x^2sin^2(y) \leq \varepsilon$ is insufficient to prove that the limit is zero. (I'm assuming that $xsin(y)$ was a typo)

Start by using the fact that $0 \leq sin^2(y) \leq 1$ and then find some function of delta $u(\delta)<\varepsilon$ such that

$$\left| \frac{x^2sin^2(y)}{x^2+2y^2} \right| \leq u(\delta)$$

 

[ZPlayer]

Thanks a lot. Check it out. Here is deal. Let me know if it good.