# Does $(f_n(x))= (\frac{nx}{1+nx^2})$ converge pointwise/uniformly on $I= [0,1]$?

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?

• You are correct. Just make sure to note all of the details: that $f_{n}(0) \to 0$, and if $x \neq 0$, $f_{n}(x) \to \dfrac{1}{x}$. Clearly, as $x \to 0$, $\dfrac{1}{x} \to \infty$, so $\lim \limits_{x \to 0} f(x) \neq f(0)$, and thus $f$ is not continuous at $x = 0$. – layman May 14 '15 at 15:18

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 $I$.