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?
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.