# Given a function $f(x,y)$ find the limit as $(x,y)\to(0,0)$

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?

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 $t,x \rightarrow 0$
You can use the inequality $$|\frac{t^2 \sin^2(x)}{2x^2+t^2}| \le \frac{t^2x^2}{x^2+t^2} \le \frac{t^2x^2}{2|xt|}=\frac{|xt|}{2}$$
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!