Use a convergence theorem to evaluate the integral I am working with the Monotone, Dominant, and Bounded Convergence Theorems. I am attempting to evaluate the integral 

lim$_{n\rightarrow\infty} \int_{0}^{1} \big(1+\displaystyle\frac{x}{n}\big)^{-n} \text{cos}\big(\frac{x}{n}\big) dx$

I have that $f_n(x) = \big(1+\displaystyle\frac{x}{n}\big)^{-n} \text{cos}\big(\frac{x}{n}\big)$ which is a continuous function on $[0,1]$ therefore it is Riemann Integrable. 

$|f_n| = |\big(1+\displaystyle\frac{x}{n}\big)^{-n} \text{cos}\big(\frac{x}{n}\big)| \leq (1+0)^{-n} \cdot \text{cos}(0) = 0$

I am stuck on the rest. Now I have to show that $f_n\rightarrow$ to something almost everywhere.
It is a standard fact that $(1+\frac x n)^{n} \to e^{x}$ as $n \to \infty$. Hence $(1+\frac x n)^{-n} \to e^{-x}$ as $n \to \infty$. Since $\cos (\frac x n ) \to 1$ we get $f_n (x) \to e^{-x}$. To apply Dominated Convergence Theorem just note that $|f_n(x)| \leq 1$ (since $1+\frac x n \geq 1$ and $|\cos \frac x n| \leq 1$) and the constant function 1 is integrable. Hence the required limit is $\int_0^{1} e^{-x}dx=1-\frac 1 e$.
Hint: Observe
\begin{align}
\left(1+\frac{x}{m} \right)^m  \leq \left(1+\frac{x}{n} \right)^n
\end{align}
whenever $m\leq n$. 


   Additional Hint: Take $m=2$. 


Here's the spoiler:
Observe we have   


   \begin{align}\left( 1+\frac{x}{n}\right)^{-n} \leq \left(1+\frac{x}{2}\right)^{-2}\end{align}


for all $n\geq 2$. Hence we have that


   \begin{align} \left|\left(1+\frac{x}{n} \right)^{-n}\cos\frac{x}{n} \right| \leq \left(1+\frac{x}{2}\right)^{-2}. \end{align}


Since $(1+x/2)^{-2}$ is integrable then by Dominated Convergence Theorem, we have that


   \begin{align} \lim_{n\rightarrow \infty}\int^1_0\left(1+\frac{x}{n}\right)^{-n}\cos\frac{x}{n}\ dx = \int^1_0 e^{-x}\ dx = 1-e^{-1}.\end{align}