<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] 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 $\frac{1}{2}+ O(\frac{1}{x})$. 
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