<s>[INST] Calculate $\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})$ I know that $$\lim\limits_{  x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})\\=\lim\limits_{  x\to + \infty}x\cdot \sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right).$$ If $x \rightarrow + \infty$, then $\sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right)\rightarrow \sin0 $. However I have also $x$ before $\sin x$ and I don't know how to calculate it.
 [/INST] \begin{align}
\lim_{x \to \infty} x \cdot \left( \sin \left( \sqrt{x^2+3}-\sqrt{x^2+2}\right)\right)&=\lim_{x \to \infty} x \cdot  \sin \left( \frac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)\\
&=\lim_{x \to \infty} \frac{x}{\sqrt{x^2+3}+\sqrt{x^2+2}}\\
&=\lim_{x \to \infty} \frac{1}{\sqrt{1+\frac{3}{x^2}}+\sqrt{1+\frac{2}{x^2}}}\\
&= \frac12
\end{align}
</s>