Limit and Integration are interchangeable

Question

Let $f_n:[0,1]\to\mathbb{R}$ be defined by $f_n(x)=\frac{n+x^3\cos x}{ne^x+x^5\sin x}$, $n\geq 1$. Find $\lim_{n\to\infty}\int_0^1f_n(x)dx$.

Approach: Here I found the limit function $f(x)=e^{-x}$. So, if I can interchange the limit and integration, then easy to find solution. But we can do that only when the convergence to $f$ is uniform. I am not getting the idea to show this convergence is uniform.

Answer 1

$ 0\leq f_n(x) \leq \frac {n+1} {ne^{x}-1}$. Now $\frac {n+1} {ne^{x}-1} \leq \frac {n+1} {n-1} \leq 2$ for $n \geq 3$ for all $x$ so we can apply DC T to interchange the limit and the integral.

Alternatively you can verify that $|f_n(x)-e^{-x}| \leq \frac {1+e} {n-1}$ which gives uniform convergence.

Answer 2

Note that for $x\in [0,1]$, $0\leq \sin(x)\leq 1$, $0\leq \cos(x)\leq 1$, $0<e^{-x}\leq 1$ and $$|f_n(x)-e^{-x}|=e^{-x}\left|\frac{1+\frac{x^3\cos x}{n}}{1+\frac{x^5e^{-x}\sin x}{n}}-1\right|=\frac{e^{-x}}{n}\frac{|x^3\cos x-x^5e^{-x}\sin x|}{1+\frac{x^5e^{-x}\sin x}{n}}\\ \leq \frac{1}{n}\frac{|x^3\cos x|+|x^5e^{-x}\sin x|}{1}\leq \frac{2}{n}.$$ and therefore $f_n\to e^{-x}$ uniformly over $[0,1]$.