Uniform and pointwise convergence of sequence of function of $f_n(x) = [\log(1+x)]^n$

I need to study the pointwise convergence of

$$f_n(x) = [\log(1+x)]^n$$

for every $$x$$ of the domain of the functions.

After i have to prove that the sequence of functions $$f_n(x)$$ is uniformly convergent to $$f$$ on the interval $$[\frac{1}{2},1]$$.

I've shown that in $$(-1,\infty)$$

$$dom(f_n(x))=\{x:x>-1\}$$

so for $$x=0$$ $$f_n(x)=0 \xrightarrow{} 0$$ for n $$\rightarrow$$ + $$\infty$$

It remains to show what happen for $$x\neq 0$$ but i don't find the pointwise convergence. How can i proceed? Thanks in advance for any help.

@andrew The pointwise limit is clearly $$0$$ for $$x\in(-1+e^{-1},e-1)$$. On $$[1/2,1]$$ you have $$|\log(x+1)^n-0|=|\log(1+x)|^n<(\log 2)^n\to 0$$ and this establishes the uniform convergence to zero.
• how do you understand that the pointwise limit is $0$ for $x \in (-1+e^{-1},e-1)$? – andrew Jan 22 at 19:00
• Only for those values you have $\log(1+x)$ is in $(-1,1)$ and increasing powers approach zero. If $x=-1+e^{-1}$ your sequence is $(-1)^n$ which has no limit. If $x=e-1$ it is constant equal to 1, so the limit is $1$ there. Outside the given interval your sequence is $A^n$ with $A\notin[-1,1]$ and is thus has no limit if $A<1$ and the limit is $+\infty$ if $A>1$. – GReyes Jan 22 at 19:54