Let $f(x,y)=\dfrac{x^a\sin(y)}{x^2+4y^2}$ $\forall (x,y)\neq (0,0)$ and $f(0,0)=0$.
Find the values of the parameter $a$ for which the function is continuous in $(0,0)$.
In the textbook the final answer was $a>1$. But I think it is not right since:
$$ 0\leq \left|\frac{x^a\sin(y)}{x^2+4y^2}\right| \leq \left|\frac{x^ay}{x^2+4y^2}\right| \leq \left|\frac{x^ay}{x^2}\right| \leq \left|x^{a-2}y\right| $$
So we get that the limit is 0 iff $a\geq 2$.
Is that fine? If we use polar cooirdinates it should lead to the same result however it gives $a>1$.
Yes, the answer is $a>1$.
If $a>1$, then\begin{align}\left|\frac{x^a\sin y}{x^2+4y^2}\right|&\leqslant\frac{|x|^{a-1}}2\frac{|x(2y)|}{x^2+(2y)^2}\\&\leqslant\frac{|x|^{a-1}}4,\end{align}since we always have $|ab|\leqslant\frac{a^2+b^2}2$. And $\lim_{(x,y)\to(0,0)}\frac{|x|^{a-1}}4=0$.
And if $a\leqslant1$,$$f(x,x)=\frac{x^{a+1}}{5x^2}\cdot\frac{\sin x}x$$and $\lim_{x\to0}\frac{x^{a+1}}{5x^2}\cdot\frac{\sin x}x$ is either $\frac15$ (if $a=1$) or $\infty$ (otherwise).