Question

Show that the sequence {fn}of functions, where


fn (x )=n/(x+2n)

is uniformly

convergent in [0.k] where k > 0.

Answer 1

Definition. If we define

$$d_n=\sup\limits_{x\in E}|f_n(x)-f(x)|$$

then $f_n$ converges to $f$  uniformly if and only if ${\displaystyle d_{n}\to 0}$  as ${\displaystyle n\to \infty }$ .

( More information: https://en.wikipedia.org/wiki/Uniform_convergence )

In our case,

1 STEP:

$$\lim\limits_{n\to\infty}f_n(x)=\lim\limits_{n\to\infty}\left(\frac{n}{x+2n}\right)=\frac{1}{2}$$

2 STEP:

$$\left|\frac{n}{x+2n}-\frac{1}{2}\right|=\left|\frac{2n-(x+2n)}{2(x+2n)}\right|=\left|\frac{-x}{2(x+2n)}\right|=\\[0.5cm] \frac{x}{2(x+2n)}$$

3 STEP: we must find $sup$ for the expression above on the interval $x\in[0;k], \forall k>0$.

To do this, we will look at the expression above as a certain function $y(x)$ and use the derivative.

$$y'(x)=\frac{d}{dx}\left(\frac{x}{2(x+2n)}\right)=\frac{1\cdot(2(x+2n))-x\cdot2}{(2(x+2n))^2}=\\[0.5cm] =\frac{4n}{4(x+2n)^2}=\frac{n}{(x+2n)^2}>0, \forall x\in[0;k],\forall n\in\mathbb{N}.$$

Then,

$$d_n=\sup\limits_{x\in[0;k]}y(x)=\sup\limits_{x\in[0;k]}\frac{x}{2(x+2n)}=\frac{k}{2(k+2n)}$$

3 STEP:

$$\lim\limits_{n\to\infty}d_n=\lim\limits_{n\to\infty}\frac{k}{2(k+2n)}=0$$

Conclusion,

$$\boxed{f_n(x)=\frac{x}{x+2n}\rightrightarrows\frac{1}{2}}$$

(the sequence is uniformly convergent).