Uniform convergence of $f_n(x) = \frac{nx}{(2+nx)(4+x^2)}$

Question

I need to study the uniform convergence of

$$f_n(x) = \frac{nx}{(2+nx)(4+x^2)}$$

on the interval $[2,+\infty)$

I've shown that on $[0,+\infty)$:

• at $x =0$ $f_n(0)=0 \xrightarrow{} 0$

• at $x \neq 0$ $$f_n(x) \xrightarrow{} \frac{x}{4x+x^3}$$

Then the $f_n(x)$ converges pointwise to : $$f(x)=\begin{cases} 0 \text{ if x=0}\\ \frac{x}{4x+x^3} \text{if x }\neq 0 \end{cases} $$ on $[0,+\infty).$ It remains to prove the uniform convergence on the interval $[2,+\infty)$.

It remains to show that: $\:$ $\underset{x\in (2,+ \infty)}{sup}|f_n(x)-f(x)| \rightarrow 0$ for n $\rightarrow + \infty $

$$sup|f_n(x)-f(x)|= sup\Big|\frac{nx}{(2+nx)(4+x^2)}-\frac{x}{4x+x^3}\Big|=\\$$.

$$= sup\Big|\frac{nx^2-2nx-x^2}{(2n+x)x(4+x^2)}\Big|=sup\Big|\frac{xn-(x+2n)}{(2n+x)(4+x^2)}\Big|\\\\$$.

I tried to compute the derivate of this quantity but i dont' find the global maximum point of $f_n$.

How can i find this point to check the uniform convergence ?

Thanks for the help in advance!!

Answer 1

First note that the pointwise limit is $\frac 1 {4+x^{2}}$. $|f_n(x)-\frac 1 {4+x^{2}}| =|\frac 2 {(4+x^{2})(2+nx)}| \leq \frac 2 {4 (2+nx)} \leq \frac 2 {8n}$ which makes it obviuous that the convergence is uniform.