Uniform convergence of sequence of functions that converges pointwise to an unbounded function I have the following sequence of functions before me:
$f_n(x)=\dfrac{1-x^n}{1+x},-1<x<1$
I have to determine whether above sequence of functions is uniformly convergent or not.
First of all, I checked pointwise convergence.
For this I used the fact that
as $n$ tends to $\infty$, $x^n$ tends to $0$ for $-1<x<1$
Hence we get that
Pointwise limit is : $f(x)=\dfrac{1}{1+x},-1<x<1$.
I,however, have reservations about the existence of above limit when $x$ approaches $-1$ from the right side.
Next I argued that $f(x)$ is an unbounded function on $-1<x<1$
as $f(x)$ tends to $\infty$ as $x$ tends to $-1$ from right side.
As a uniformly convergent sequence of functions must converge to a function which is continuous and hence bounded, $f_n(x)$ does not converge uniformly.
Is this line of thinking correct? Please suggest.
 A: Since the pointwise limit $f(x) = 1/(1+x)$ is continuous on $(-1,1)$ your strategy to disprove uniform convergence will not work without arguing that the uniform convergence of the $f_n$ on $(-1,1)$ is equivalent to uniform convergence on $(-1,1]$ (since then the pointwise limit will be discontinuous at the right endpoint). Yet to prove this extension property you have to work with explicit estimates of the sup norm, so we might as well check the definition of uniform convergence instead.
Recall $f_n\to f$ uniformly on $(-1,1)$ when $\lim_n\|f_n-f\|_\infty=0$, where $$\|f-f\|_\infty = \sup_{x\in(-1,1)}|f_n(x)-f(x)|.$$ We have
$$|f_n(x) - f(x)| = \left|\frac{1-x^{n}}{1+x} - \frac{1}{1+x}\right| = \frac{|x|^{n}}{|1+x|}.$$
Since $|1+x|\leq 1+|x|\leq 2$, we have $|f_n(x)-f(x)|\geq |x|^{n}/2$. Yet for each $n$ the sup of the right-hand side is $1/2$. Thus $$\|f_n-f\|_\infty\geq 1/2$$ and the sequence cannot converge uniformly (since if it did converge uniformly, the uniform limit and the pointwise limit would coincide).
