Find the value of :Solving $\lim_{n \to \infty} \sqrt{n} \sin\left({\sqrt{n+3}-\sqrt{n-2}}\right)$ I have trouble finding the value of the following limit:
$$\lim_{n \to \infty} \sqrt{n} \sin\left({\sqrt{n+3}-\sqrt{n-2}}\right)$$
For now I have rewritten the term into:
$$ \lim_{n \to \infty} \dfrac{\sin\left({\sqrt{n+3}-\sqrt{n-2}}\right)}{\large \frac{1}{\sqrt{n}}}$$
Now I have a limit of type $\large \frac{0}{0}$ so I think I could use L'Hopital's rule. But I would like to know if there is a way you can solve this without using L'Hopital's rule. 
Thanks for your answers.
 A: $\sqrt{n + 3} - \sqrt{n - 2} = \frac{5}{\sqrt{n + 3} + \sqrt{n - 2}}$
This yields the following term:
$$\sqrt{n}\sin(\sqrt{n + 3} - \sqrt{n - 2}) = \frac{5 \sqrt{n}}{\sqrt{n + 3} + \sqrt{n - 2}} \frac{\sin\left(\frac{5}{\sqrt{n + 3} + \sqrt{n - 2}}\right)}{\frac{5}{\sqrt{n + 3} + \sqrt{n - 2}}}$$
Now use the fact that $\frac{\sin(h)}{h} \to 1$ for $h \to 0$ and $\frac{5 \sqrt{n}}{\sqrt{n + 3} + \sqrt{n - 2}} = \frac{5}{\sqrt{1 + 3/n} + \sqrt{1 - 2/n}} \to \frac{5}{2}$.
A: With equivalents:
$$\sin(\sqrt{n+3}-\sqrt{n-2})=\sin\biggl(\frac5{\sqrt{n+3}+\sqrt{n-2}}\biggr)\sim_\infty\frac5{\sqrt{n+3}+\sqrt{n-2}}\sim_\infty\frac5{2\sqrt n},$$
hence
$$\sqrt n\sin(\sqrt{n+3}-\sqrt{n-2})\sim_\infty\sqrt n\,\frac5{2\sqrt n}=\frac52.$$
A: Hint: conjugation leads you to: $L = \displaystyle \lim_{n \to \infty} \dfrac{5\sqrt{n}}{\sqrt{n+3}+\sqrt{n-2}} = \dfrac{5}{2}$.
A: Notice,  $$\lim_{n\to \infty}\sqrt n\sin\left(\sqrt{n+3}-\sqrt{n-2}\right)$$
$$=\lim_{n\to \infty}\underbrace{\left(\frac{\sin\left(\sqrt{n+3}-\sqrt{n-2}\right)}{\sqrt{n+3}-\sqrt{n-2}}\right)}_{\longrightarrow 1}\cdot \sqrt n(\sqrt{n+3}-\sqrt{n-2})$$
$$=\lim_{n\to \infty}\sqrt n(\sqrt{n+3}-\sqrt{n-2})$$
$$=\lim_{n\to \infty} \frac{\sqrt n(\sqrt{n+3}-\sqrt{n-2})(\sqrt{n+3}+\sqrt{n-2})}{(\sqrt{n+3}+\sqrt{n-2})}$$
$$=\lim_{n\to \infty} \frac{5\sqrt n}{(\sqrt{n+3}+\sqrt{n-2})}$$
$$=5\lim_{n\to \infty} \frac{1}{\sqrt{1+\frac3n}+\sqrt{1-\frac2n}}$$
$$=5\cdot \frac{1}{1+1}=\color{red}{\frac 52}$$
