# Bivariate Limits - is my solution correct?

$\require{cancel}$

Find for which values of $\alpha\in\mathbb{R^+}$ the following function is continuous in $\mathbb{R^2}$.

## $$f(x,y)=\begin{cases} \frac{x^2+y^2}{|x|^{\alpha}y}\sin\left(\frac{|x|^{\alpha}y}{x^2+y^2}\right), & \text{if x\neq0 and y\neq0} \\[2ex] 1, & \text{if x=0 or y=0} \end{cases}$$

I first started by evaluating the limit, $$\lim_{(x,y)\to (0,0)}\frac{|x|^{\alpha}y}{x^2+y^2}\to \textstyle \text{Polar Substitution} \to \displaystyle \lim_{r\to 0}\frac{|r|^{\alpha}\cdot\left|\cos(\theta)\right|^{\alpha}\cdot \cancel{r}\sin(\theta)}{r^\cancel{2}}= \\ = \displaystyle \lim_{r\to 0}\left(|r|^{\alpha-1}\cdot\left|\cos(\theta)\right|^{\alpha}\sin(\theta)\right)=\begin{cases} 0, & \text{if \alpha>1} \\[2ex] \text{Indeterminate}, & \text{if \alpha\leq1} \end{cases}$$ And used this result to compute the limit, \lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{|x|^{\alpha}y}\sin\left(\frac{|x|^{\alpha}y}{x^2+y^2}\right)\to \textstyle \begin{align} &u=\frac{|x|^{\alpha}y}{x^2+y^2} \\ &u\to0\text{ as }(x,y)\to(0,0) \end{align} \to \displaystyle \lim_{u\to 0}\frac{\sin(u)}{u}=1 which can only be done if $\alpha>1$.

Therefore $f(x,y)$ is continuous $\{\forall\alpha\in\mathbb{R^+}\mid \alpha>1\}$.

Yes, it is correct. However you may express finding the limit $\sin(u)/u$ for $u\to0$ and reducing the question to checking for which $\alpha$ your $u$ tends to $0$.