multivariable limit question

Question

Is this an acceptable solution? $$\lim_{(x,y)\rightarrow(0,0)}\frac{\sin(2(x^2+y^2))}{x^2+y^2}$$ $$t=x^2+y^2$$ So $t\rightarrow0$. Now I change the limit to: $$\lim_{t\rightarrow0}\frac{\sin(2t)}{t}=2$$ This solution makes sense to me, but I am still having doubts.

Answer 1

Your proof is fine. In fact, it's very similar to the common technique of switching to polar coordinates. In that technique, one puts $x = r\cos\theta$ and $y = r\sin\theta$ and evaluates the limit as $r \to 0$. This is equivalent to putting $r^2 = x^2 + y^2$. You've put $r = x^2 + y^2$ instead. Check out this question for more information on this technique and others.

Answer 2

Or, indeed, the technique used for evaluating $$\lim_{x\to 0}\frac{\sin ax}{x}.$$ Any time $u\to 0$ as $x\to 0$, it is easy to check with the official limit definition that $$\lim_{x\to 0} f(u(x)) = \lim_{u\to 0} f(u)\,.$$ The same applies when $x$ is a vector.