Limits of $ \mathop {\lim }\limits_{n \to + \infty } \left( {1 + n + n\cos n} \right)^{\frac{1}{{2n + n\sin n}}} $ 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

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 $e^0=1$.
