Simple limit in multi variable For $x=(x_1,x_2,x_3)$, determine the limit $$\lim_{x\to 0} \frac{\sin|x|^2}{|x|^2+x_1x_2x_3}. $$
I want to use that $\lim_{x\to 0} \frac{sin|x|^2}{|x|^2} = 1$ but I can't see how to do that. Any hints? 
 A: Let , $$ f(x)=\frac{\sin|x|^2}{|x|^2+x_1x_2x_3}.$$
So , $$\frac{1}{f(x)}=\frac{|x|^2+x_1x_2x_3}{\sin|x|^2}=\frac{|x|^2}{\sin|x|^2}+\frac{|x|^2}{\sin|x|^2}\cdot\frac{x_1x_2x_3}{|x|^2},$$
which implies
$$\lim_{x\to 0}\frac{1}{f(x)}=1+\lim_{x\to 0}\frac{x_1x_2x_3}{|x|^2}.$$
Now we calculate the limit ,  $\lim_{x\to 0}\frac{x_1x_2x_3}{|x|^2}.$
Let, $\epsilon >0$ and choose $\delta_i>0$ such that $\frac{\epsilon}{\sqrt 3}<\delta_i$ for , $i=1,2,3$.
For this, put $x_1=r\sin \theta \cos \phi$ , $x_2=r\sin \theta \sin \phi$ , $x_3=r\cos \theta$. Then , $$\left|\frac{x_1x_2x_3}{|x|^2}\right|=\left|\frac{r^3\sin^2\theta\cos \theta\sin \phi \cos \phi}{ r^2}\right|\le r<\epsilon $$whenever, $r<\epsilon$ , i.e. $r^2<\epsilon^2$ , i.e. $x_1^2+x_2^2+x_3^2<\epsilon^2$ , i.e. $x_i^2<\frac{\epsilon^2}{3}(say)$ , i.e. $|x_i|<\frac{\epsilon}{\sqrt 3}<\delta_i$
Choose , $\delta=\min\{\delta_1,\delta_2,\delta_3\}$.
Then , $\left|\frac{x_1x_2x_3}{|x|^2}-0\right|<\epsilon$ , whenever $|x_i-0|<\delta$
So, $\lim_{x\to 0}\frac{x_1x_2x_3}{|x|^2}=0.$
Hence, $\lim_{x\to 0}f(x)=1$
