Limit involving delta-epsilon proof

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SUMMARY

The limit of the function f(x,y) = x^2 * sin^2(y) / (x^2 + 2y^2) as (x,y) approaches (0,0) is proven to be 0 using the delta-epsilon method. The key condition to satisfy is |f(x,y) - 0| ≤ ε. It is established that 0 ≤ sin^2(y) ≤ 1, which aids in bounding the function. A suitable function u(δ) must be found such that |f(x,y)| ≤ u(δ) < ε to complete the proof.

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ZPlayer
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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.
 
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What have you tried so far?
 
What I tried so far is here.
2yni3no.jpg
 
ZPlayer said:
What I tried so far is here.
2yni3no.jpg

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)&lt;\varepsilon such that

\left| \frac{x^2sin^2(y)}{x^2+2y^2} \right| \leq u(\delta)
 
Thanks a lot. Check it out. Here is deal. Let me know if it good.
2rcy7n9.jpg
 

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