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To calculate the limts

the First

$\mathop {\lim }\limits_{x \to 0^ + } \left( {2\sin \sqrt x + \sqrt x \sin \frac{1}{x}} \right)^x$

Suppose:

$\frac{1}{x} = n$ We find

$ \mathop {\lim }\limits_{x \to 0^ + } \left( {2\sin \sqrt x + \sqrt x \sin \frac{1}{x}} \right)^x = \mathop {\lim }\limits_{n \to + \infty } \left( {2\sin \frac{1}{{n^2 }} + \frac{1}{{n^2 }}\sin n} \right)^{\frac{1}{n}} = 0 $

The second

$\mathop {\lim }\limits_{n \to + \infty } \left( {1 + n + n\cos n} \right)^{\frac{1}{{2n + n\sin n}}}$ No idea to me

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For $n$ sufficiently large:

$$0\le\frac{1}{2n+n\sin n}\log(1+n+n\cos n)\le\frac{\log(3n)}{n}=\frac{\log3+\log n}{n}\xrightarrow{n\to\infty}0$$ hence the desired limit is $e^0=1$.

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