Jtdylktuy

$(f_n)$$_n$$_in $$_mathbb N$ is a sequence of functions where $f_n : [0,2pi] to mathbb R$ $ forall n in mathbb N$. Find all values of $x in [0,2pi]$ such that $(f_n)$$_n$$_in $$_mathbb N$ converges and find pointwise limit if it exists.

(i) $f_n (x) = $sin$ (frac{x}{n})$

(ii)$f_n (x) = $sin$ (nx)$

(iii)$f_n (x) = $sin$^n$ $(x)$

(i) This is easy to show: $lim_{n to infty} f_n(x) = f(x)$ where $f(x) = 0 forall x in [0,2pi]$

So $f_n(x)$ converges $forall x in [0,2pi]$

(ii) $f_n(x)$ converges if $x = 0, pi, 2pi$. And the pointwise limit is
$f(x) = 0$ if $x = 0, pi, 2pi$

(iii)
$$f_n(x) = begin{cases} 0, & text{if $x = 0, pi, 2pi$} \[2ex] sin^n (x), &
text{if $xneq 0, pi, 2pi$} end{cases}$$

Then we have

$$lim_{n to infty} f_n(x) = f(x) = begin{cases} 0, & text{if $x neq frac{3pi}{2} $} \[2ex] 1, &
text{if $x = frac{pi}{2}$} \[2ex] Doesn't exist, &
text{if $x = frac{3pi}{2}$} end{cases}$$

So $f_n(x)$ converges if $x in [0, 2pi]$ {$frac{3pi}{2}$}

$$$$

I somewhat feel that my answers for (ii) and (iii) are wrong.

Any comment / correction is appreciated

I do not know if you need to prove your answers. If that is the case, there are many steps missing. In any case, (ii) is correct. (iii) is basically an exponential, for that to converge the base belongs to $(-1,1]$, in your case $-1<sin(x)leq1$ which implies $xneq3pi/2$, and convergence is as you said.