prove that $\frac{1-e^{-x^2}}{x}\le 2\sqrt{2} , \ x>0$, Can you show very easy methods? I hope I'll see many methods. Thank you everyone.
Prove that:
$$\frac{1-e^{-x^2}}{x}\le 2\sqrt{2} \ \ \ \qquad \forall  x>0.$$
 A: It's clear for $x > 1$, the value is bounded above by $1$ since $1 - e^{-x^2} < 1$
If $0 < x \leq 1$ on the other hand, write
$$e^{-x^2} = 1 - x^2 + \dfrac{x^4}{2!} - \dots $$
and so $$\dfrac{1 - e^{-x^2}}{x} = x - \dfrac{x^3}{2!} + \dfrac{x^5}{3!} - \dots $$
This is an alternating series with decreasing coefficients since $x^n < x^m$ for $n > m$.  Therefore it is bounded by the first term and so $$\dfrac{1 - e^{-x^2}}{x} < 1$$
for all $x$.  (Thanks to @Andre for pointing out the simplification with the alternating series)
A: We shall prove the stronger inequality $\frac{1-e^{-x^2}}{x} \leq \sqrt{e-1}<2\sqrt{2}$.
This is equivalent to 
$$1 \leq x\sqrt{e-1}+e^{-x^2}$$
If $x \geq \frac{1}{\sqrt{e-1}}$, then $$x\sqrt{e-1}+e^{-x^2} \geq 1+e^{-x^2} \geq 0$$.
Otherwise $x<\frac{1}{\sqrt{e-1}}$, then by Bernoulli's inequality $e^{-x^2} \geq 1-(e-1)x^2$, so $$x\sqrt{e-1}+e^{-x^2} \geq x\sqrt{e-1}+1-(e-1)x^2=1+x(\sqrt{e-1}-(e-1)x) \geq 1$$
where the last inequality holds since $\sqrt{e-1}-(e-1)x>\sqrt{e-1}-(e-1)\frac{1}{\sqrt{e-1}}=0$.
A: $$
1-e^{-x^2}=\int_0^x2te^{-t^2}dt\;\Rightarrow\;\frac{1-e^{-x^2}}{x}=2\int_0^x\frac{t}{x}e^{-t^2}dt\leq 2 \int_0^xe^{-t^2}dt\leq 2\int_0^{+\infty}e^{-t^2}dt=\sqrt{\pi}
$$
A: You can also use the Mean Value theorem.
First notice that if $x \lt 0$ or $x \gt 1$ it is easy to show the inequality.
So suppose $0 \lt x \le 1$.
Let $f(x) = -e^{-x^2}$ (note: there is a minus sign there).
Then we have that $f'(x) = 2x e^{-x^2}$, which attains a maximum $x = \frac{1}{\sqrt{2}}$ (we can check using the second derivative $= 2e^{-x^2}(1 - 2x^2)$).
By the mean value theorem, we have that there is some $\eta \in (0,x)$ such that
$$\frac{1 - e^{-x^2}}{x} = \frac{f(x) - f(0)}{x} = f'(\eta) \le f'\left(\frac{1}{\sqrt{2}}\right) =  \sqrt{\frac{2}{e}}$$
