Answer to Question #197916 in Real Analysis for Nazam farooq

Question

Define uniform convergence of a sequence of functions. Discuss the uniform convergence of the sequence nx^2+1/nx+1 the interval  [1. 2]

Answer 1

In a hyper-real context, uniform convergence allows for a simpler description. So, if is infinitely close to for all x in the space of and all infinite m, a sequence converges to f uniformly.(For a related concept of uniform continuity, see micro continuity)

$f(x)=\frac{nx^2+1}{nx+1}$

$\lim\limits_{x \to \infin}f_n(x)=\lim\limits_{x \to \infin}[\frac{nx^2+1}{nx+1}]=\lim\limits_{x \to \infin}[\frac{\frac{n}{n}x^2+\frac{1}{n}}{\frac{nx}{n}+\frac{1}{n}}]=\lim\limits_{x \to \infin}[\frac{x^2+\frac{1}{n}}{x+\frac{1}{n}}]=\lim\limits_{x \to \infin}[\frac{x^2+\frac{1}{\infin}}{x+\frac{1}{\infin}}]= x$

When x=0

$f_n(x)=0$

Therefore, $f(x)={x \space if \space x \not =0\brack 0 \space if \space x=0}$

$\lim\limits_{x \to \infin}f_n(x)= f(x)$ for all x.

To prove that the above convergence is not pointwise

Suppose that $f_n \to f$ uniformly

Therefore, $\exist \varepsilon >0, \exist x>[1,2], \forall N \epsilon \N, \exist n> N$

$|f_n(x)-f(x)|< \varepsilon$

Now , taking $\varepsilon = \frac{1}{10}, x=\frac{1}{\sqrt{n}}$ and n=N+1

Therefore, $|\frac{nx^2+1}{nx+1}-x|=\frac{1-x}{nx+1}>\varepsilon=\frac{1}{10}$

Which is a contradiction to the fact that $f_n(x)$ converges uniformly

Hence , $f_n(x)$ does not converge to f uniformly.