Does $\displaystyle(f_n(x))= \bigg(\frac{nx}{1+nx^2}\bigg)$ converge pointwise/uniformly on $I= [0,1]$?
My attempt:
Pointwise:
$\displaystyle \lim_{n \to \infty}f_n(x) = \lim_{n \to \infty} \frac{x}{\frac{1}{n}+ x^2} = \frac{1}{x}$.
However, notice that $f(x) = \frac{1}{x}$ is discontinuous at the point $x=0$, hence $(f_n(x))$ does not converge pointwise on $I = [0,1]$. Consequently it does not converge uniformly on $I = [0,1]$.
Is my reasoning correct? Or am I missing something?
Best Answer
What you have done to evaluate the limit of $f_n(x)$ only works if $x>0$ since the rule $\lim \frac{a_n}{b_n} = \frac{\lim a_n}{\lim b_n}$ requires the right hand side limits to exist and $\lim b_n \neq 0$ which is not the case here for $x=0$. So you have to treat $x=0$ seperately. This however is easy since $f_n(0) = 0$ for all $n$. So $\lim_{n\to\infty}f_n(x)$ exists for all $x\in I$ and the sequence is pointwise convergent.
But, as you already pointed out, the limit is not continuous in $x=0$. If the sequence would converge uniformly though, this had to be the case since all $f_n$ are continuous. Thus, the sequence doesn't converge uniformly on $I$.