Uniform convergence of sequence of functions $\frac{2+nx^2}{2+nx}$ on [0,1]?

Question

I have recently been trying some questions related to the uniform convergence of a sequence of functions. And meanwhile, I got stuck in one of the problems in which I have been supposed to discuss the point-wise and uniform convergence of the sequence of functions defined as $$f_n(x)=\frac{2+nx^2}{2+nx}$$ on the interval $[0,1]$

I have found out its point-wise limit that is given by $$f(x)= \begin{cases} 1, \text {if } x = 0,1\\ x, \text {if } 0 < x < 1 \end{cases}$$ So the first half has been done.

In the second half, let $\epsilon>0$ be given.

Now I need to find an $m$(if possible) such that $|f_n(x)-f(x)| <\epsilon$ for all $x$ in $[0,1]$ and for all $n\geq m.$ So I see that if $x$ is $0$ or $1$, then any natural number $m$ will work.

But the problem is when $x$ is neither of them.

If I assume that we get a natural number $m$ such that the definition of uniform convergence holds then or further calculations, we get that $n>\dfrac{2(1-x-\epsilon)}{x\epsilon}$ for all $n\geq m.$

Now I am not understanding how to proceed.

Help, please!

Answer 1

To prove using just the definition that that the convergence is not uniform note that $|f_n(x)-f(x)|=|\frac {2-2x} {2+nx}|$ for $0<x<1$. Put $x=\frac 1 n$ to get $\frac {2-2/n} {2+1} \to \frac 2 3$. This shows that $\sup_x |f_n(x)-f(x)|$ does not tend to $0$.

Answer 2

Since $f$ is discontinuous, so the convergence can't be uniform.

Answer 3

As the $f_n$ are continuous on the compact interval $[0,1]$, if the convergence was uniform, the limit $f$ would also be continuous. And it is not as $f(0) = 1$ while $f(x) = x$ for $x \neq 0$.

A contradiction proving that the convergence can't be uniform.

Answer 4

Observe that the sequence $(|f_n(1/n)-f(1/n)|)$ does not converge to $0$ as $ n \to \infty.$

Hence, the convergence can't be uniform.