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Let $\beta >0$, how to prove that for all $a\in \textbf{R}^n$,

$$\lim_{x\rightarrow 0} \frac{x\cdot a}{|x|^{1+\beta}} = \lim_{x\rightarrow 0} \frac{x_1a_1+...+x_na_n}{(\sqrt{x_1^2+...+x_n^2})^{1+\beta}} $$ exists?

I know in the real line, $\lim_{x\rightarrow0}\frac{ax}{|x|}$ DNE, is this related to the case in $\textbf{R}^n$? Or, is it helpful to convert this question into polar coordinate or modulus?

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    The limit does not exist for constant $a$ unless $a=0.$ If $a\ne 0$ and $x=a/m$ then $x\cdot a/\|x\|^{1+b}=\|a\|^{1-b}m^b$ which $\to \infty$ as $m\to \infty.$ On the other hand if $x=-a/m$ then $x\cdot a/\|x\|^{1+b}=-\|a\|^{1-b}m^b$ goes to $-\infty$ as $m\to \infty.$ – 2017-02-26

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For any real number $r$, we set the following function $$sgn(r)=\begin{cases}1, &r>0\\ -1, &r\leq 0\end{cases}.$$ And for $a\in\mathbb{R}^n$, we define $$\text{sgn}(a)=(sgn(a_1),\dots,sgn(a_n)).$$ Then we observe that $$\text{sgn}(a)\cdot a=|a_1|+\dots+|a_n|.$$ If $a=0$ then obviously the limit is $0$. If not, let $$x_\epsilon=\epsilon \text{sgn}(a),\quad x_\epsilon'=-\epsilon\text{sgn}(a)$$ where $\epsilon>0$. Then $$\lim_{\epsilon\to 0^+}\frac{x_\epsilon\cdot a}{|x_\epsilon|^{1+\beta}}=\lim_{\epsilon\to 0^+}\frac{\epsilon(|a_1|+\dots+|a_n|)}{(\epsilon^2 n)^{\frac{1+\beta}{2}}}=+\infty,$$ $$\lim_{\epsilon\to 0^+}\frac{x_\epsilon'\cdot a}{|x_\epsilon'|^{1+\beta}}=\lim_{\epsilon\to 0^+}\frac{-\epsilon(|a_1|+\dots+|a_n|)}{(\epsilon^2 n)^{\frac{1+\beta}{2}}}=-\infty.$$ Thus we have different limits in two different directions, and this means the limit does not exist.