$|x-\frac{1}{2}|^{1+1/n}\rightarrow |x-\frac{1}{2}|$? I want to show that $C^1[0,1]$ isn't a Banach Space with the norm:
$$||f||=\max\limits_{y\in[0,1]}|f(y)|$$
Therefore, I want to show that the sequence $\left \{ |x-\frac{1}{2}|^{1+\frac{1}{n}} \right \}$ converges to $|x-\frac{1}{2}|$, but I can't find $N$ in the definition of convergence. Could anyone give me an idea?
Thank You. 
 A: You have, using the Mean Value Theorem,   the inequality
$$
1-e^t\leq |t|,\ \ \ \ t\leq0.
$$
Then, for $t\in[0,1]$ and $x>0$,
$$
|t^{1+x}-t|=|t|\,|t^x-1|\leq |t|\,|e^{x\log t}-1|\leq x\,|t\log t|\leq \frac xe.
$$
So
$$
\left|\,\left|x-\tfrac12\right|^{1+\tfrac1n}-\left| x-\tfrac12\right|\,\right |\leq\frac1{ne}
$$
A: To show $\left|x-\frac12\right|^{1+1/n}=\left|x-\frac12\right|\left|x-\frac12\right|^{1/n}\to \left|x-\frac12\right|$ pointwise, which is clear when $x=\frac12$, it suffices to show that when $0<t<1$, $t^{1/n}\to 1$.  Note that $(t^{1/n})$ is bounded above by $1$ and monotone increasing, so it has a limit $L$, with $L>0$.  The subsequence $(t^{1/(2n)})$ must converge to the same limit $L$, and the sequences of squares of this subsequence must converge to $L^2$, implying that $L^2=L$.  Hence $L=1$.
From the above, we have monotone pointwise convergence of the sequence to $\left|x-\frac12\right|$.  By Dini's Theorem, the convergence is uniform, hence convergence in the given norm.
