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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.

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  • $\begingroup$ In my opinion, L'Hopital's Rule would be optimal here. $\endgroup$ – K. Jiang Feb 7 '16 at 22:11
  • $\begingroup$ this is not an indeterminant form of the type 0/0. Hint: $\sqrt{n+3}-\sqrt{n-2}=\frac{5}{\sqrt{n+3}+\sqrt{n-2}}$ $\endgroup$ – Matematleta Feb 7 '16 at 22:21
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$\sqrt{n + 3} - \sqrt{n - 2} = \frac{5}{\sqrt{n + 3} + \sqrt{n - 2}}$

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 $\frac{5 \sqrt{n}}{\sqrt{n + 3} + \sqrt{n - 2}} = \frac{5}{\sqrt{1 + 3/n} + \sqrt{1 - 2/n}} \to \frac{5}{2}$.

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  • $\begingroup$ In you line after "following term", in the left side the square root of $\;n\;$ must be multiplying, not dividing, that sine. $\endgroup$ – DonAntonio Feb 7 '16 at 23:43
  • $\begingroup$ @Joanpemo Thank's for pointing that out, I've fixed it now. $\endgroup$ – Dominik Feb 7 '16 at 23:48
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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 $$\sqrt n\sin(\sqrt{n+3}-\sqrt{n-2})\sim_\infty\sqrt n\,\frac5{2\sqrt n}=\frac52.$$

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Hint: conjugation leads you to: $L = \displaystyle \lim_{n \to \infty} \dfrac{5\sqrt{n}}{\sqrt{n+3}+\sqrt{n-2}} = \dfrac{5}{2}$.

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Notice, $$\lim_{n\to \infty}\sqrt n\sin\left(\sqrt{n+3}-\sqrt{n-2}\right)$$ $$=\lim_{n\to \infty}\underbrace{\left(\frac{\sin\left(\sqrt{n+3}-\sqrt{n-2}\right)}{\sqrt{n+3}-\sqrt{n-2}}\right)}_{\longrightarrow 1}\cdot \sqrt n(\sqrt{n+3}-\sqrt{n-2})$$ $$=\lim_{n\to \infty}\sqrt n(\sqrt{n+3}-\sqrt{n-2})$$ $$=\lim_{n\to \infty} \frac{\sqrt n(\sqrt{n+3}-\sqrt{n-2})(\sqrt{n+3}+\sqrt{n-2})}{(\sqrt{n+3}+\sqrt{n-2})}$$ $$=\lim_{n\to \infty} \frac{5\sqrt n}{(\sqrt{n+3}+\sqrt{n-2})}$$ $$=5\lim_{n\to \infty} \frac{1}{\sqrt{1+\frac3n}+\sqrt{1-\frac2n}}$$ $$=5\cdot \frac{1}{1+1}=\color{red}{\frac 52}$$

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