Prove that $\left|\frac{e^{-ht}-1}{h}\right|\le t$ for $h>0$ Prove that $$\left|\frac{e^{-ht}-1}{h}\right|\le t$$ for $h>0$, $t>0$.
$$\left|\frac{e^{-ht}-1}{h}\right|=\left|\frac{1+(-ht)+\frac{(-ht)^2}{2!}+\frac{(-ht)^3}{3!}+\ldots -1}{h}\right|=$$
$$=\left|-t+\frac{ht^2}{2}+\frac{-h^2 t^3}{3!}+\ldots \right|=\left|\sum_{n=1}^{\infty} (-1)^n \frac{h^{n-1}t^n}{n!} \right|$$
Now, since this is alternating series and 
$$\frac{h^{n-1}t^n}{n!}$$
is non-increasing sequence we have that
$$\left|\sum_{n=1}^{\infty} (-1)^n \frac{h^{n-1}t^n}{n!} \right|\le|a_1|=|-t|=t$$
Is the above thing correct?
 A: For fixed $t$, let $f:[0,\infty)\to\mathbb{R}$ given by $f(h)=e^{-ht}$. Fixing $h>0$, by the Mean Value Theorem there is $x\in(0,h)$ such that $f^\prime(x)=\frac{e^{-ht}-1}{h}$. That is $\frac{e^{-ht}-1}{h}=-te^{-xt}$.
Then $$\left|\frac{e^{-ht}-1}{h}\right|=|te^{-xt}|=te^{-xt}\le te^{-0t}=t$$
A: Just to add a second method:
$$\left\vert \frac{e^{-ht}-1}{h}\right\vert=\left\vert\int_0^te^{-hx}dx\right\vert\le \int_0^tdt=t$$
A: Here, we present a way forward that does not rely on the MVT.  In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequality
$$\bbox[5px,border:2px solid #C0A000]{e^x\ge 1+x} \tag 1$$
for $x>-1$.  And since $e^x>0$ for all $x$, then $(1)$ is true for $x\le-1$ also.
Then, from $(1)$, we have 
$$e^{-x}\ge 1-x \tag 2$$
for all $x$.  Rearranging $(2)$ reveals that
$$1-e^{-x}\le x \tag 3$$
Setting $x=ht$ in $(3)$ and dividing by $h>0$ yields
$$\frac{1-e^{ht}}{h}\le t \tag 4$$
whereupon noting both sides of $(4)$ are non-negative for $ht>0$, we obtain the coveted inequality
$$\bbox[5px,border:2px solid #C0A000]{\left|\frac{1-e^{ht}}{h}\right|\le t} \tag 5$$
as was to be shown!  

The Only Tool Used Here Was The Inequality $(1)$, 
  which relied on only $(i)$ The limit definition of the exponential function; and $(ii)$ Bernoulli's Inequality

