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If it exists, find

$$\lim_{(t,x)\to(0,0)}\frac{t^2\sin^2(x)}{2x^2+t^2}$$

Along the curves $x=mt,t=0,x=at^2,t=ax^2$ the limit approaches 0; the graph also makes $L=0$ seem correct.

So assuming that $L=0$, I start the epsilon delta proof: $$0<\sqrt{x^2+t^2}<\delta$$ $$\left|\frac{t^2\sin^2(x)}{2x^2+t^2}\right|<\epsilon$$

All attempts I did trying to find a $\delta$ for every $\epsilon$ have just end up circling around and accomplishing nothing. How am I supposed to complete this proof?

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Use the Squeeze Theorem. Notice that $t^{2} < t^{2} + 2x^{2}$, so $$ \frac{t^{2}}{t^{2} + 2x^{2}} \leq 1$$ Then we have that $$ 0 \leq \left| \frac{t^{2}\sin^{2}{x}}{2x^{2} + t^{2}} \right| \leq |\sin^{2}{x}|$$ and you can apply the Squeeze Theorem accordingly, noting how $|\sin^{2}{x}|$ behaves as $t,x \rightarrow 0$

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You can use the inequality $$|\frac{t^2 \sin^2(x)}{2x^2+t^2}| \le \frac{t^2x^2}{x^2+t^2} \le \frac{t^2x^2}{2|xt|}=\frac{|xt|}{2}$$

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Given $\epsilon >0$ we have

$$\begin{align} \left|\frac{t^2\sin^2 x}{2x^2+t^2}\right|&\le \frac{t^2\,x^2}{x^2+t^2}\\\\ &\le \frac{(x^2+t^2)^2}{x^2+t^2}\\\\ &=x^2+t^2\\\\ &<\epsilon \end{align}$$

whenever $\sqrt{x^2+t^2}<\delta=\sqrt{\epsilon}$. And we are done!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. If the answer was not useful to warrant an up vote, I am happy to delete. $\endgroup$ – Mark Viola Oct 21 '15 at 23:14

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