Show that $f_n(x)$ converges uniformly. Show that $f_n(x) = \frac {n+\cos(x)}{ne^x+\sin(x)}$ converges uniformly.   
I think it converges uniformly to $\frac {1}{e^x}.$ But, I am struggling to prove this with the formal definition.  
Attempt: $\forall\varepsilon>0, \exists M\in N$ such that for $n\ge M,$ $$\left|\frac {n+\cos(x)}{ne^x+\sin(x)}-\frac {1}{e^x}\right|=\left|\frac {1+\frac{\cos(x)}{n}}{e^x+\frac{\sin(x)}{n}}-\frac{1}{e^x}\right|\to\left|\frac{1}{e^x}-\frac{1}{e^x}\right|=0<\varepsilon$$ for all $x \in R.$  Is this okay? Is there another way to show this? 
Edit: The original question is:  
Consider the sequence of functions $$f_n(x) = \frac {n+\cos(x)}{ne^x+\sin(x)}, n=1,2,...$$ Compute the following limit of integrals $$\lim_{n \to \infty}\int_0^1 \frac {n+\cos(x)}{ne^x+\sin(x)}dx.$$ 
I was trying to solve this question by using the theorem that if $f_n$ converges uniformly to $f$, $\lim_{n \to \infty} \int f_n = \int f.$ If $f_n$ does not converges uniformly to $f$, how can we solve this question?
 A: In fact the convergence is only pointwise but not uniform. To see this, you can pick a sequence of $x$, say $x_n:=-2n\pi$, then 
$$|f_n(x_n)-e^{-x_n}|=\frac1ne^{2n\pi}$$
Which blows up and clearly doesn't converge to zero. And hence you should see that there exists an $\epsilon>0$ (in fact all of them, in this particular case) such that whatever $M\in\Bbb N$ you choose, you can always find some $n>M$ such that
$\sup_x |f_n(x)-e^{-x}|\ge\epsilon.$
What does uniform convergence mean, intuitively? You can picture it this way: if we have $f_n$ converges to $f$ uniformly on some interval $I$ (the specification of the domain of convergence is extremely important for us to decide whether the convergence is uniform or not), it means that given any $\epsilon>0$, if you draw a $\pm\epsilon$ vertical band around the graph of $f$ over $I$ then as $n$ goes past some point ($M$) the graph of $f_n$ will stay within that band henceforth forever. 
A: $f_n$ converges pointwise to $f$ in $\mathbb R$.
However, $f_n$ converges UNIFORMLY to $f$ in every interval of the form $[x_0,\infty)$, where $x_0\in\mathbb R$. 
