Is $M=\{x(t)\in C_{[0,1]}:x(t)=e^{-\alpha t}\}$ pre-compact? 
Study the pre compactness of $M=\{x(t)\in C_{[0,1]}:x(t)=e^{-\alpha t}\}$, for $\alpha\in\mathbb{R}$.

I am supposed to use the Arzela-Ascoli theorem:

Theorem:For a set $M$ to be pre-compact it is necessary and sufficient  this set to be equicontinuous. 

So I tried to prove that $\exists \delta>0,\epsilon>0,\forall t_1,t_2\in[0,1] $
$$|t_1-t_2|<\delta\:\:\:\:\:\:\:|x(t_1)-x(t_2)|<\epsilon\:\:\:\forall x(t)\in M$$
So I shall try to prove: 
$\exists \delta>0,\epsilon>0,\forall t_1,t_2\in[0,1] $
$$|t_1-t_2|<\delta\:\:\:\:\:\:\:|e^{-\alpha t_1}-e^{-\alpha t_2}|<\epsilon\:\:\:\forall x(t)\in M$$
Using Lagrange Mean Value theorem:
$\frac{dx(t)}{dt}=-\alpha e^{-\alpha t}$
For $\xi\in [0,1]$
$|e^{-\alpha t_1}-e^{-\alpha t_2}|=-\alpha e^{-\alpha\xi}|t_1-t_2|\leqslant?$
Question:
1) I do not know how to proceed. How should I do it?
2) Is there another easier method to major $|e^{-\alpha t_1}-e^{-\alpha t_2}|$?
Thanks in advance!

So it is not equicontinuous. Suppose that it were, there is some $\delta>0$ such that for all $t_{1},t_{2}\in[0,1]$ with $|t_{1}-t_{2}|\leq\delta$ and $\alpha\in{\bf{R}}$, we have $|e^{-\alpha t_{1}}-e^{-\alpha t_{2}}|<1$. Now take $t_{1}=\delta$, $t_{2}=0$, we have $|e^{-\alpha\delta}-1|<1$ and hence $e^{-\alpha\delta}<2$ for all $\alpha\in{\bf{R}}$. Taking $\alpha\rightarrow-\infty$, the expression blows up.
