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.

The ordo calculus is very useful for such problems. Sometimes you do not need the full power of Taylor series, only the first couple of coefficients. 
$\sqrt{x^2+3}- \sqrt{x^2+2} = x(\sqrt{1+3/x^2}- \sqrt{1+2/x^2})=$ 
$= x(1+\frac{1}{2}\cdot(3/x^2)+O(1/x^4)- 1-\frac{1}{2}\cdot(2/x^2)+O(1/x^4))= \frac{1}{2x}+O(1/x^3)$. 
Now $\sin(\frac{1}{2x}+O(1/x^3))= \frac{1}{2x}+O(1/x^2)$ as $x\rightarrow \infty$, thus the expression is $\frac{1}{2}+ O(\frac{1}{x})$. 
