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.
 A: \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}
A: Letting $h=\frac1x$:
$$\begin{array}{cl}
&\displaystyle \lim_{x \to \infty} x \sin \left( \sqrt{x^2+3} - \sqrt{x^2+2} \right) \\
=&\displaystyle \lim_{x \to \infty} x \sin \left( \frac 1 {\sqrt{x^2+3} + \sqrt{x^2+2} } \right) \\
=&\displaystyle \lim_{h \to 0^+} \frac1h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\
=&\displaystyle \lim_{h \to 0^+} \frac 1 {\sqrt{1+3h^2} + \sqrt{1+2h^2}} \frac {\sqrt{1+3h^2} + \sqrt{1+2h^2}} h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\
=&\displaystyle \frac12 \times \lim_{h \to 0^+} \frac {\sqrt{1+3h^2} + \sqrt{1+2h^2}} h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\
=&\displaystyle \frac12 \times 1 \\
=&\dfrac12
\end{array}$$
A: 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 $\frac{1}{2}+ O(\frac{1}{x})$. 
A: \begin{align}
   x \sin \left( \sqrt{x^2+3}-\sqrt{x^2+2}\right)
   &=x  \sin \left( \frac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)\\
   &=\left(\dfrac{x}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)
     \dfrac{\sin \left( \dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}
            {\left(\dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}\\
   &=\left(\dfrac{1}{\sqrt{1+\frac{3}{x^2}}+\sqrt{1+\frac{2}{x^2}}}\right)
     \dfrac{\sin \left( \dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}
            {\left(\dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}\\
   &\to \dfrac 12 \cdot 1  \ \text{as $x \to \infty$}\\
   &\to \dfrac 12 \ \text{as $x \to \infty$}
\end{align}
