real analysis - Calculate $limlimits_{ xto + infty}xcdot sin(sqrt{x^{2}+3}-sqrt{x^{2}+2})$

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.

Answer

Letting $h=\frac1x$:

$$\begin{array}{cl} &\displaystyle \lim_{x \to \infty} x \sin \left( \sqrt{x^2+3} - \sqrt{x^2+2} \right) \\ =&\displaystyle \lim_{x \to \infty} x \sin \left( \frac 1 {\sqrt{x^2+3} + \sqrt{x^2+2} } \right) \\ =&\displaystyle \lim_{h \to 0^+} \frac1h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\ =&\displaystyle \lim_{h \to 0^+} \frac 1 {\sqrt{1+3h^2} + \sqrt{1+2h^2}} \frac {\sqrt{1+3h^2} + \sqrt{1+2h^2}} h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\ =&\displaystyle \frac12 \times \lim_{h \to 0^+} \frac {\sqrt{1+3h^2} + \sqrt{1+2h^2}} h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\ =&\displaystyle \frac12 \times 1 \\ =&\dfrac12 \end{array}$$