Finding Multivariable limits using polar coordinates

Question

How do I find the limit of this multivariable function as it goes to zero using polar coordinates?

$$ \frac{\sin (x^2 + y^2)}{(x^2 + y^2)^2} $$

Answer 1

Use \begin{align} x &= r \cos \theta \\ y &= r \sin \theta \end{align} So $x^2 + y^2 = r^2$ hence \begin{equation} \frac{\sin (x^2 + y^2)}{(x^2 + y^2)^2} = \frac{\sin r^2}{r^4} \end{equation} Using L'Hopital twice, we get \begin{equation} \frac{\sin r^2}{r^4} \sim \frac{2\cos\left(r^2\right)-4r^2\sin\left(r^2\right)}{12r^2} \rightarrow +\infty \end{equation}

Answer 2

We can avoid polar coordinates in that case indeed

$$(x,y) \to 0 \iff t=x^2+y^2 \to 0^+$$

therefore by standard limit

$$\frac{\sin (x^2 + y^2)}{(x^2 + y^2)^2}=\frac{\sin t}{t^2}=\frac 1t \cdot \frac{\sin t}t\to \infty $$