For which values of $x \in I$ is $(f_n)$ differentiable term by term? 
Let $f_n(x) = \frac{1}{n}e^{-nx}$ on $x\in[0,1] =I$.
  
  
*
  
*Discuss the pointwise and uniform convergence of $(f_n)$ on $I$.
  
*For which values of $x \in I$ is $(f_n)$ differentiable term by term?
  

My attempt:

1.
Pointwise Convergence:
$\displaystyle \lim_{n \to \infty} f_n(x) = 0$ for all $x \in I$. 
Hence $(f_n)$ converges pointwise to $f(x)=0$ on $I$.
Uniform Convergence:
$\|f_n-0\| = \sup \bigg\{\bigg|\frac{1}{n}e^{-nx} \bigg| : x \in [0,1] \bigg\} = \bigg| \frac{1}{n} \bigg| = \frac{1}{n} \to 0$ as $n \to \infty$.
Hence $(f_n)$ converges uniformly to $f(x) = 0$ on $I$.

Is this correct?

2.
This is the question which I am stuck at.
I know that, in order for $(f_n)$ to be term by term differentiable
  
  
*
  
*$(f_n)$ must converge pointwise
  
*$f'_n$ must be continuous and 
  
*$(f'_n)$ must converge uniformly. 
However, this does not occur on the entire interval $I = [0,1]$. Can anyone please show me how I can go about finding the the values of $x$ in $I$ for which this is true?


$f'_n(x)=e^{-nx}$ is a geometric series that converges uniformly on $[a,1]$ for $0<a<1$. It follows that $\{f_n\}$ is differentiable term by term on $(0,1]$.
