<s>[INST] Limit Using squeeze theorem Find the limit of the following function as $x \rightarrow 0$
$$
\frac{|x|}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}} \sin \left(\frac{1}{3 \sqrt{x}}\right)
$$
$\lim _{x \rightarrow 0} \frac{|x|}{\sqrt{\left(x^{4}+4 x^{2}+77\right)}} \sin \left(\frac{1}{3 \sqrt{x}}\right)$
My approach,
applying squeeze theorem,
$
-1 \leq \sin \left(\frac{1}{3 \sqrt{x}}\right) \leq 1
$
$-\frac{|x|}{\sqrt{x^{2}+4 x^{2}+7}} \leq \frac{|x|}{\sqrt{x^{4}+4 x^{2}+7}} \sin \left(\frac{1}{\sqrt{3x}}\right) \leq \frac{|x|}{\sqrt{x^{4}+4 x^{2}+7}}$
$\operatorname{Now,}_{\operatorname{limit}_{x \rightarrow 0} \frac{-x}{\sqrt{x^{4}+4 x^{2}+7}}}=\ _{x \rightarrow 0} \frac{x}{\sqrt{x^{4}+4 x^{2}+7}}=0$
Hence answer is zero. 
Am I correct?
 [/INST] Other approach:
Let $f(x)=\frac{|x|}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}} \sin \left(\frac{1}{3 \sqrt{x}}\right)$
We note that $f(x)=-f(-x)$, hence $f$ is odd, we can study it for $x > 0$.
Remembering that
$$
\lim_{x \to 0} \frac{\sin x}{x} = 1
$$
we get
$$
\begin{split}
\lim_{x \to 0^+} \frac{x}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}} \sin \left(\frac{1}{3 \sqrt{x}}\right) &= \lim_{x \to 0^+} \frac{x}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}} \frac{\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\frac{1}{3 \sqrt{x}}} \frac{1}{3 \sqrt{x}} \\
&=\lim_{x \to 0^+} \frac{x^{1-\frac{1}{2}}}{3} \cdot \lim_{x \to 0^+}  \frac{\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\frac{1}{3 \sqrt{x}}}\cdot \lim_{x \to 0^+} \frac{1}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}}  \\
&=\lim_{x \to 0^+} \frac{\sqrt x}{3} \cdot 1 \cdot \frac{1}{\sqrt 7} = 0 \cdot 1 \cdot \frac{1}{\sqrt 7}= 0
\end{split} 
$$
</s>