# Study the continuity domain and derivability domain

Let $(u_n)_{n\in\mathbb{N}}$ with the general term $u_n=\frac {1+x^{n}}{1+x+x^{2}+...+x^{n+p-1}}$, where $x\ge0$ and $p \in \mathbb{N}$. Let $f(x)= \lim_{n\to\infty}u_n$. Find the differentiability and continuity domain.

First I tried to simplify a little $u_n$ using the sum of the geometric progression and I got this $$u_n=\frac{(1+x^{n})(x-1)}{x^{n+p}-1}.$$ So if $x \in (0,1)$, then $f(x)=1-x$.

What should I do when $x = 1$ and $x>1$?

First of all, I think you have a slight error in you the sum of a geometric progression formula: the denominator must be $x^{n+p}-1$, not $x^{n+p\color{red}{-1}}-1$.
If $x=1$, you can literally plug it into the original expression for $u_n$.
If $x>1$, then the let's divide the numerator and denominator by $x^n$: $$u_n=\frac{(1+x^n)(x-1)}{x^{n+p}-1}=\frac{\left(\frac{1}{x^n}+1\right)(x-1)}{x^p-\frac{1}{x^n}},$$ and observe that in this case $1/x^n\to0$ as $n\to\infty$.
• If I plug in 1 in the original expression doesn't it interfere some errors? $\frac 00$? – C. Cristi Apr 18 '18 at 13:47
• @C.Cristi: Not into the simplified, but into the original: $u_n(x)=\frac{1+x^n}{1+x+x^2+...+x^{n+p-1}}$, so $u_n(1)=\cdots$. – zipirovich Apr 18 '18 at 13:50
• Yeah, right. If I'm right $u_n(1)=0$? – C. Cristi Apr 18 '18 at 13:50