# Calculus limit pointwise and uniform convergence

Consider the sequence $(f_n)_{n\in\mathbb{N}}$ of functions $f_n : [0, \pi]\to\mathbb{R}$ defined by $f_n(x) = \sin^n(x)$. Show that:

1) The sequence $(f_n)_n$ converges pointwise, find its pointwise limit 2) The sequence $(f_n)_n$ does not converge uniformly

For the first point I know that I have to compute the limit for $n \to\infty$ of $f_n(x)$, so: $\lim_{n \to\infty}\inf (\sin^n(x))$. How to compute this limit, knowing that $x$ belongs to the interval $[0, \pi]$ ?

Hint. We have that $f_n(\pi/2)=1$. Moreover $f_n(x)$ is the power of a number in $[0,1)$ for $x\in [0,\pi/2)\cup(\pi/2,\pi]$. So, what is the limit function $f$?

As regards the second question, note that the limit function $f$ is not continuous in $[0,\pi]$.

Hints:

(1) $\sin(x)\ge 0$ and can be $= 1$ or $<1$.

(2) What property must have the uniform limit if exists?

If $x=\frac {\pi}{2}$, then $f_n (x)=1$ and $$\lim_{n\to+\infty}f_n (x)=1$$

If $x\neq \frac {\pi}{2}$ then $0\leq \sin (x)<1$ and $$\lim_{n\to+\infty}\sin^n (x)=0$$

the pointwise limit $f$ is defined by:

$$f (x )=0 \, \text{if}\; x\neq \frac {\pi}{2}$$ and $f (\frac {\pi}{2})=1$.

As said by @Robert, the functions $f_n$ are continuous at $[0,\pi]$ and $f$ discontinuous at $[0,\pi]$, so the convergence is not uniform at $[0,\pi]$.

I think we've beat the pointwise convergence issue to death, but just wanted to add that the limit function is not continuous implies $(f_n)_n$ certainly does not converge uniformly relies on the assumption that uniform continuity implies continuity (this can easily be proven).

If you don't want to prove that, you can also just apply the definition of uniform continuity directly (which is the same idea, but more explicit). Given any $x, y \in [0, \pi]$, does it hold that $\exists$ $\delta$ st $\lvert x - y \rvert < \delta$ $\implies$ $\lvert f(x) - f(y) \rvert < \epsilon$ for any choice of $\epsilon > 0$? Hint: choose $\epsilon$ to be any number less than $1$ and $y = \frac{\pi}{2}$ and see what happens.