I read a similar question that was solved by using the fact that $|\sin(x)| \leq |x|$, but I tried a different approach, as I struggle with utilizing inequalities to solve limits of this kind. Please note that my textbook did not ask for the limit to be solved without using inequalities.
I attempted to prove this result by first switching to polar coordinates, then applying Hopital's rule as follows:
$$\lim_{(x,y) \to (0,0)} \frac{\sin^2 xy}{x^2 + y^2}=\lim_{\rho \to 0} \frac{\sin^2 (\rho^2\sin(\theta)\cos(\theta))}{\rho^2}$$
Now I use cosine duplication formula to rewrite $\sin^2(t)$
$$\lim_{\rho \to 0} \frac{\sin^2 (\rho^2\sin(\theta)\cos(\theta))}{\rho^2}=\lim_{\rho \to 0} \biggl(\frac{1}{2}\biggr)\frac{1-\cos(2\rho^2\sin(\theta)\cos(\theta)}{\rho^2}$$
and now I apply Hopital's rule
$$\lim_{\rho \to 0} \biggl(\frac{1}{2}\biggr)\frac{4\rho\sin(\theta)\cos(\theta)\sin(2\rho^2\sin(\theta)\cos(\theta))}{2\rho}=\lim_{\rho \to 0} \sin(\theta)\cos(\theta)\sin(2\rho^2\sin(\theta)\cos(\theta))$$
Now if I'm understanding this correctly, because the argument of the second sine function goes to zero, the result is proven.
Have I made any mistakes? Was there a faster or more intuitive approach to solving this problem without using inequalities?
Best Answer
Another way
First rewrite the expression: $$ \frac{\sin^2 xy}{x^2 + y^2} = \left( \frac{\sin xy}{xy} \right)^2 \frac{x^2y^2}{x^2+y^2} $$
Here the first factor tends to $1$ since $\frac{\sin t}{t} \to 1$ as $t \to 0$ and $xy \to 0$ as $(x,y) \to (0,0).$
For the second factor we use polar coordinates: $$ \frac{x^2y^2}{x^2+y^2} = \frac{\rho^4 \cos^2\theta \sin^2\theta}{\rho^2} = \rho^2 \cos^2\theta \sin^2\theta \to 0 $$ as $\rho \to 0$ which is the case when $(x,y) \to (0,0).$
Thus, $$ \frac{\sin^2 xy}{x^2 + y^2} \to 1^2 \cdot 0 = 0. $$