Showing that a sequence is Cauchy but not convergent Consider $X$ to be the set of all continuous real-valued functions on $C[0,1]$ with metric 
$$d(x,y) = \int_{0}^{1} |x(t) - y(t)| dt$$
Show that $x_n(t) = \left\{\begin{matrix}
 n& 0 \leq t \leq n^{-2} \\ 
 t^{-1/2}& n^{-2} \leq t \leq 1
\end{matrix}\right.$ is Cauchy, but does not converge.
So here is what I did, I basically assumed that $m > n$, (for some $m$) $$d(x_n, x_m) = \int_{0}^{m^{-2}}|x_n-x_m|dt + \int_{m^{-2}}^{n^{-2}}|x_n-x_m|dt + \int_{n^{-2}}^{1}|x_n-x_m|dt = \int_{0}^{m^{-2}}|n-m|dt + 0 + 0 = (m - n)m^{-2}$$
Now according to the book, the answer should've been $m^{-1} - n^{-1}$ for $n > m$. 
As for convergence, I got $d(x_n,x) = \int_{0}^{n^{-2}} |x_n - x| dt + \int_{n^{-2}}^{1} |t^{-1/2} - x| dt \to 0$ implies that $x(t) =1/\sqrt{t}$ on $[0,1]$. But $x(0)$ is undefined, so not continuous. 
I am not sure if I did this right, I have more confidence in the convergence part over the Cauchy part.
 A: If $m>n$, you have $d(x_m,x_n) = \int_{0}^\frac{1}{m^2} |m-n| dt + \int_{\frac{1}{m^2}}^\frac{1}{n^2} |\frac{1}{\sqrt{t}}-n| dt + \int_{\frac{1}{n^2}}^1 |\frac{1}{\sqrt{t}}-\frac{1}{\sqrt{t}}| dt = \frac{m-n}{mn}$.
 It follows that the sequence is Cauchy.
You have to do a little more work to show that the sequence $x_n$ does not converge to a continuous function. Your idea is right, but you need to show that $d(x,x_n) \to 0$ implies that $x_n(t) \to \frac{1}{\sqrt{t}}$ on $(0,1]$. Another way is to show directly that $x_n$ cannot converge to any continuous function $f$. As you observed, it hinges on the behavior at $t=0$.
Let $f$ be a continuous function. Then for some $k>0$, and $\epsilon>0$, we have $f(t) \leq k$ for $t \in [0,\epsilon]$. Without loss of generality, we may choose $\epsilon < \frac{1}{k^2}$. Let $n > \frac{2}{\sqrt{\epsilon}}$, then
\begin{eqnarray}
d(x_n,f) &\geq& \int_{[0,\epsilon]} (x_n(t)-k) dt \\
&=& \int_{[0,\frac{1}{n^2}]}(n-k)dt +\int_{[\frac{1}{n^2},\epsilon]}(\frac{1}{\sqrt{t}}-k)dt \\
&=& \frac{1}{n^2}(n-k)+2\sqrt{\epsilon}-k\epsilon+\frac{k}{n^2}-\frac{2}{n} \\
&=& 2\sqrt{\epsilon}-k\epsilon-\frac{1}{n} \\
&=& \sqrt{\epsilon}(2-k \sqrt{\epsilon})-\frac{1}{n} \\
&\geq & \sqrt{\epsilon} -\frac{1}{n} \\
& = & \frac{\sqrt{\epsilon}}{2} > 0
\end{eqnarray}
It follows that $x_n$ cannot converge to any continuous function.
