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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?

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Lozansky
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1 Answers1

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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, $\lim_{x\to 0}f(x)=1$

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  • How did you find $\delta$? – Lozansky May 12 '15 at 20:27
  • How does $x_1^2+x_2^2+x_3^2 < \epsilon^2$ imply $x_i^2 < \frac{\epsilon^2}{3}$? Is $x_i$ in this case the largest element from the set and so $x_i^2+x_i^2+x_i^2=3x_i^2$ is the largest possible sum (i.e. the upper restriction)? – Lozansky May 13 '15 at 12:17
  • We assume it.....$x_1^2<\frac{\epsilon^2}{3}$ , $x_2^2<\frac{\epsilon^2}{3}$ , $x_3^2<\frac{\epsilon^2}{3}$ so that $x_1^2+x_2^2+x_3^2<\epsilon$. – Empty May 13 '15 at 15:20
  • You may take anything$>0$ instead of $\frac{\epsilon^2}{3}$ so that their sum is $<\epsilon^2$..For example , you may take , $x_1^2<\frac{\epsilon^2}{2}$ , $x_2^2<\frac{\epsilon^2}{4}$ , $x_3^2<\frac{\epsilon^2}{4}$ such that $x_1^2+x_2^2+x_3^2<\epsilon^2$. – Empty May 14 '15 at 03:18