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Let{$ {a_n} $} be sequence of real numbers s.t $ |a_n| \le \sqrt{n} , \,\, n=1,2,3\ldots$

Find

$$ \lim_{n \to \infty} e^{\frac{a_n}{n}} + \sqrt{n}\sin(\frac{\sqrt{2}}{\sqrt{n}})$$

My attempt got me limit of sin part in question which most probably is $\sqrt{2}$ but what to do with exponential part?

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    $\begingroup$ Note $|a_n|/n\rightarrow0$. $\endgroup$ – David Mitra Jan 11 '15 at 16:24
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Since $|a_n/n| \le 1/\sqrt{n} \to 0$ as $n\to \infty$, $\lim_{n\to \infty} e^{a_n/n} = e^0 = 1$. Now $$\lim_{n\to \infty} \sqrt{n}\sin \frac{\sqrt{2}}{\sqrt{n}} = \lim_{n\to \infty} \sqrt{2} \frac{\sin \sqrt{\frac{2}{n}}}{\sqrt{\frac{2}{n}}} = \lim_{x\to 0^+} \sqrt{2}\frac{\sin x}{x} = \sqrt{2}\lim_{x\to 0^+} \frac{\sin x}{x} = \sqrt{2}.$$

Therefore, your limit is $1 + \sqrt{2}$.

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Hint $$ \frac{a_n}{n}\leq \frac{\sqrt{n}}{n}= \frac{1}{\sqrt{n}}$$

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  • $\begingroup$ So do we get upper bound 1 of limit for the exponential part ? $\endgroup$ – ketan Jan 11 '15 at 16:54
  • $\begingroup$ Is the answer 1 + $\sqrt{2}$? $\endgroup$ – ketan Jan 11 '15 at 17:04

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