Proving that $(C^1, {\left\lVert \cdot\right\rVert}_\infty)$ is not complete through the example $f_n(x)=|x|^{1+\frac{1}{n}}$ Given the function sequence $f_n(x)=|x|^{1+\frac{1}{n}}, x\in [-1,1],\, n \geq 1, \,$ in the lecture notes it is shown that the normed vector space $(C^1, {\left\lVert \cdot\right\rVert}_\infty)$ is not complete. There are some points which I dont understand.
"$f_n$ is $C^1$ on the interval $[-1,1] \setminus \{0\},$ with the dervivative $f_n^\prime =|x|^{\frac{1}{n}} $for $x \geq 0,$ and the opposite for $x<0.$" This is the first thing I dont understand. For me: $$ f_n^\prime (x)= \begin{cases}
(1+ \frac{1}{n})x^{\frac{1}{n}}, x\geq 0 \\
 (1+ \frac{1}{n}) (-x)^{\frac{1}{n}}, x<0    \
\end{cases} $$
Can somebody give a comment if i am right ?
"Since $f(x)=o(|x|), \,f$ is differentiable at $0$, with derivative equal to $0$." This the second point I dont understand, particularly the expression  $f(x)=o(|x|)$, since it is clear that: $$f(x)= \,\text{lim} |x|^{1+\frac{1}{n}}=|x|= \begin{cases}
x+o(x), x\geq 0 \\
 -x + o(x), x<0    \
\end{cases}$$
Am I wrong ? I only used the Taylor expension of first order around the point $0.$
And last, I dont understand the last inequality of the following expression:$$|f_n(x)-f(x)|=|x||e^{\frac{log(|x|)}{n}}-1|=|x|(1- e^{\frac{log(|x|)}{n}})\leq \frac{-|x|log(|x|)}{n}\leq \frac{1}{2n}, \, x\in[-1,1]\setminus \{0\}.$$
I will highly appreciate your comment.
 A: Your computation of $f_n'$ is correct. The statement  $f(x)=o(|x|)$ is false, as you have observed. For the inequality $\frac {-x| \log |x|} n \leq \frac 1 {2n}$ proceed as follows: the function $\log t -\frac t 2 $ defined on $[1,\infty)$ has maximum at $t=2$ and since the value of the function at 2  is negative (because $e >2$) we get $\log t -\frac t 2 <0$ for all $t \geq 1$. Put $t=\frac 1 {|x|}$ to get to get the inequality $\frac {-x| \log |x|} n \leq \frac 1 {2n}$.
A: Here is a more visually appealing example, $f_n(x) = \sqrt{x^2+{1 \over n}}$. Each $f_n$ is smooth, and
clearly $f_n(x) \to f(x) = |x|$.
Since $f_n(x)-f(x) = { 1\over n} {1 \over \sqrt{x^2+{1 \over n}} + |x|} \le {1 \over \sqrt{n}}$, we have $\|f-f_n\|_\infty \to 0$.
Another way to see the result is to note that the polynomials are dense
(in the $\|\cdot \|_\infty$ norm) in $C[-1,1]$, this follows from the 
Stone Weierstrass theorem. Since $C[-1,1]$ contains non differentiable
functions, we can obtain the same result.
