Limit epsilon delta proof

Question

How to prove the following with epsilon delta method $$\lim_{(x,y)\to (4,\pi)} x^{2} \sin \frac{y}{8} = 8 \sqrt{2}$$

P.S. This question is already asked in this forum but it has no solution. Please provide an idea on how to go about this problem

Answer 1

In the product $f(x)⋅g(y)$, of two continuous functions $f,g$, for any given $ϵ<1$ select $δ$ so that for $|x-x_0|<δ$, $|y-y_0|<δ$ you get $|f(x)-f(x_0)|<ϵ/M$, $|g(y)-g(y_0)|<ϵ/M$ with some $M\ge1$ that will depend on $x_0,y_0$ and $ϵ$. Then \begin{align} |f(x)⋅g(y)-f(x_0)⋅g(y_0)| &\le |f(x)|\cdot |g(y)-g(y_0)|+|f(x)-f(x_0)|\cdot|g(y_0)|\\ &< (|f(x_0)|+ϵ+|g(y_0)|)\cdotϵ/M \end{align} Thus with selecting $M=\max(1,|f(x_0)|+|g(y_0)|+ϵ)$ this inequality chain will indeed satisfy the continuity definition.

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This proves that $$ \lim_{(x,y)\to(x_0,y_0)}f(x)g(y)=f(x_0)g(y_0). $$ To complete the task, you now only need to (confirm that the square and sine are continuous and) insert the limit point to evaluate $$ 4^2\sin\frac\pi8=8\sqrt{2-\sqrt2}. $$