Showing uniform convergence of $f_n(x) = x+\frac{x^n}{n}$ and convergence of $f_n'$ I need to show that
$$f_n(x) = x+\frac{x^n}{n}$$
converges uniformly to a differentiable function in $[0,1]$, the sequence of derivatives $f_n'$ converges pointwise in $[0,1]$, but $f_n' \neq (\lim f_n)'$
First of all, I think the limit of $f_n(x)$ is $0$, so I tried to analyze
$$\left|x+\frac{x^n}{n}\right| = \left|\frac{nx+x^n}{n}\right|<^*\left|\frac{n}{n}\right|$$
*when $x\in [0,1]$
but that doesn't help
The sequence of derivatives is $1+x^{n-1}$ which we know converges pointwise, and $[\lim f_n]' = x' = 1$ so at least I showed that $f_n' \neq [\lim f_n]'$. What about the uniform convergence of $f_n$?
 A: For any $\;x\in[0,1]\;$ :
$$x+\frac{x^n}n\xrightarrow[n\to\infty]{}x+0=x=:f(x)$$
and the convergence is uniform because for any $\;x\in[0,1]\;$ :
$$|f_n(x)-f(x)|=\frac{x^n}n\le\frac1n\xrightarrow[n\to\infty]{}0$$
Clearly, for all $\;x\in[0,1]\;:\;\;f'(x)=1\;,\;\;f_n'(x)=1+x^{n-1}\;$ , yet when $\;x=1\;$ we have that
$$f_n'(1)=1+1^{n-1}=2\xrightarrow[n\to\infty]{}2\neq1$$
A: Just observe that $\vert f_n(x)-x\vert =x^n/n \le 1/n$ so $f_n(x)\to x$ uniformly. 
A: if $f_n$ converges uniformly, then $\sup f_n(x) - f(x)| < \epsilon$
if you want to use the derivatives you have calculated then:
let $g_n = |f_n(x) - f(x)|$
Use the mean value theorem
for and $a,b \in [0,1]$ there is a $c \in (0,1)$ such that $g_n'(c) = \frac {g_n(b) - g_n(a)}{b-a}$
$\sup g(x) < g'(x)$
Now show that $n>N ,g'(x) = \epsilon$
but it is not really necessary.
I think you can demonstrate $\sup |f_n(x) - f(x)| < \epsilon$ outright. 
A: Hint for the uniform convergence if $(f_n)$ at $[0,1]$.
The pointwise limit is $x$ at $[0,1]$
and
$$\forall x\in[0,1]\; |f_n(x)-x|=|\frac{x^n}{n}|\leq\frac{1}{n}$$
You can conclude.
