Calculate $\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})$

Question

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.

Answer 1

\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}