How to Show that the Error Function has an Upper Bound? Here is the error function: 

$$\mathrm{erf}(x)=\frac{2}{\sqrt\pi}\int^x_0e^{-t^2} dt$$

Here is the question:

Show that the odd function erf is bounded, by using the fact
  that:$$e^{-t^2} \le te^{-t^2} , t \ge1$$ Remark: The normalisation of
  erf is chosen so that: $$ \lim_{x→∞}\mathrm{erf}(x) = 1.$$

I think that in order to show that the error function is bounded I need to do a substitution into the error function, however I am not sure how to go from there. 

We have immediately:
$$erf(x)=\frac{2}{\sqrt\pi}\int^x_0e^{-t^2} dt \le \frac{2}{\sqrt\pi}\left[\int_0^1 1 dt +\int^x_1 te^{-t^2} dt\right]=\frac{2}{\sqrt\pi}\left[1-\frac{1}{2}\int_1^x \frac{d}{dt}(e^{-t^2})dt\right]$$
By the fundamental theorem of calculus, the last part is equal to:
$$\frac{2}{\sqrt\pi}\left[1 -\frac{1}{2}\left(e^{-x^2} - e^{-(1^2)} \right) \right] \le \frac{2}{\sqrt\pi}\left[1 -\frac{1}{2}\left(0 - 1 \right) \right] \le \frac{3}{\sqrt{\pi}}=constant$$
This concludes the proof.
