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How do I find the given limit?

It seems that the limit without the $()^{\sqrt{x}}$ is 1, but I don't know how to solve it with the $()^{\sqrt{x}}$, i.e. $\lim _\limits{{x\to \infty }}\left(1+\sqrt{x}\sin \frac{1}{x}\right)^{\sqrt{x}}$.

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    $\begingroup$ When it comes to functions like ${f(x)}^{g(x)}$, where $f,g$ are non-constant function, taking $\log$ is very good idea. $\endgroup$ – MAN-MADE Jun 27 '17 at 9:05
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As $\lim_{x\to\infty}\sqrt x\sin\dfrac1x=\lim_{x\to\infty}\dfrac1{\sqrt x}\cdot\dfrac{\sin\dfrac1x}{\dfrac1x}\to0\cdot1$

$$\lim _\limits{{x\to \infty }}\left(1+\sqrt{x}\sin \frac{1}{x}\right)^{\sqrt{x}}=\left(\lim_{x\to\infty}\left(1+\sqrt x\sin\dfrac1x\right)^{\dfrac1{\sqrt x\sin\frac1x}}\right)^{\lim_{x\to\infty}\dfrac{\sin\dfrac1x}{\frac1x}}$$

If we set $\dfrac1{\sqrt x\sin\frac1x}=n, $ the inner limit becomes $$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$$

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Another answer: rewrite the limit as $$ \lim_{x\to \infty}\left(1+x\sin(1/x^2)\right)^{x} $$ and consider that $\sin(z)=z+O(z^3)$ as $z\to 0$. Hence $$ \left(1+\frac{1}{x}+O\left(\frac{1}{x^5}\right)\right)^{x} \to e. $$

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The expression is $$ \mathrm{exp}\left(\frac{\log(1+\sqrt{x}\sin(1/x))}{x^{-1/2}}\right). $$ Apply de l'Hopital at the exponent $$ \lim_{x\to \infty} \frac{\frac{1}{2}x^{-1/2}\sin(1/x)-x^{-3/2}\cos(1/x)}{-\frac{1}{2}x^{-3/2}(1+\sqrt{x}\sin(1/x))}=\lim_{x\to \infty}\frac{x\sin(1/x)-2\cos(1/x)}{-(1+\sqrt{x}\sin(1/x))}=1. $$ Hence, by the continuity of the exponential, the limit is $e$.

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