Limit question involving the the greatest integer function and fractional part The value of $$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\sin \left\{ {\frac{2}{n}} \right\}}}{{\left[ {2n\tan \frac{1}{n}} \right]\left( {\tan \frac{1}{n}} \right)}} + \frac{1}{{{n^2} + \cos n}}} \right)^{{n^2}}},$$ where $[.]$ denotes greatest integer function and $\{.\}$ denotes fractional part function, is _________.
My approach is as follow $\mathop {\lim }\limits_{n \to \infty } \sin \left\{ {\frac{2}{n}} \right\} = {0^ + };\mathop {\lim }\limits_{n \to \infty } \left[ {2n\tan \frac{1}{n}} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {2\frac{{\tan \frac{1}{n}}}{{\frac{1}{n}}}} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {{2^ + }} \right] = 2$
$\mathop {\lim }\limits_{n \to \infty } \tan \frac{1}{n} = 0;\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2} + \cos n}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{n\left( {1 + \frac{{\cos n}}{n}} \right)}} = 0.$
Cannot approach from here.

Recall the Maclaurin series of $\sin x$ and $\tan x$:
$$\sin x= x-\frac{x^3}6+O(x^5),\qquad x\to0,$$
$$\tan x= x+\frac{x^3}3+O(x^5),\qquad x\to0.$$
As a result,
$$\sin\left\{\frac 2n\right\}=\sin\left(\frac2n\right)=\frac2n-\frac4{3n^3}+O\left(\frac1{n^5}\right),\qquad n\to\infty,$$
$$\tan\left(\frac1n\right)=\frac1n+\frac1{3n^3}+O\left(\frac1{n^5}\right),\qquad n\to\infty.$$
Since
$$2n\tan\left(\frac1n\right)=2+\frac2{3n^2}+O\left(\frac1{n^4}\right),\qquad n\to\infty,$$
we know $\left[2n\tan\left(\frac1n\right)\right]=2$ for $n$ sufficiently large. Therefore, as $n\to\infty$, we have
\begin{align*}
\frac{{\sin \left\{ {\frac{2}{n}} \right\}}}{{\left[ {2n\tan \frac{1}{n}} \right]\left( {\tan \frac{1}{n}} \right)}}&=\frac{\frac2n-\frac4{3n^3}+O\left(\frac1{n^5}\right)}{\frac2n+\frac2{3n^3}+O\left(\frac1{n^5}\right)}\\
&=\frac{1-\frac2{3n^2}+O\left(\frac1{n^4}\right)}{1+\frac1{3n^2}+O\left(\frac1{n^4}\right)}\\
&=\left(1-\frac2{3n^2}+O\left(\frac1{n^4}\right)\right)\left(1-\frac1{3n^2}+O\left(\frac1{n^4}\right)\right)\\
&=1-\frac1{n^2}+O\left(\frac1{n^4}\right).
\end{align*}
Now, using $\log(1+x)\sim x$ as $x\to 0$ gives that, as $n\to\infty$
\begin{align*}
n^2\log\left( {\frac{{\sin \left\{ {\frac{2}{n}} \right\}}}{{\left[ {2n\tan \frac{1}{n}} \right]\left( {\tan \frac{1}{n}} \right)}} + \frac{1}{{{n^2} + \cos n}}} \right)&=n^2\log\left(1-\frac1{n^2}+\frac1{n^2+\cos n}+O\left(\frac1{n^4}\right)\right)\\
&=n^2\left(-\frac1{n^2}+\frac1{n^2+\cos n}+O\left(\frac1{n^4}\right)\right)\\
&=-1+\frac{n^2}{n^2+\cos n}+O\left(\frac1{n^2}\right);
\end{align*}
Since $\frac{n^2}{n^2+1}\leq\frac{n^2}{n^2+\cos n}\leq \frac{n^2}{n^2-1}$, by squeezing we have $\lim_{n\to\infty}\frac{n^2}{n^2+\cos n}=1$, and thus
$$\lim_{n\to\infty}n^2\log\left( {\frac{{\sin \left\{ {\frac{2}{n}} \right\}}}{{\left[ {2n\tan \frac{1}{n}} \right]\left( {\tan \frac{1}{n}} \right)}} + \frac{1}{{{n^2} + \cos n}}} \right)=0.$$
Therefore,
$$\lim_{n\to\infty}{\left( {\frac{{\sin \left\{ {\frac{2}{n}} \right\}}}{{\left[ {2n\tan \frac{1}{n}} \right]\left( {\tan \frac{1}{n}} \right)}} + \frac{1}{{{n^2} + \cos n}}} \right)^{{n^2}}}=1.$$
