Applying the Dominated Convergence Theorem In a question I've got, I found that I need to prove the following equality:
$\lim_{h\rightarrow 0}\int_{0}^{\infty}\frac{e^{-(t+h)x}-e^{-tx}}{hx}dx=\int_{0}^{\infty}\lim_{h\rightarrow 0}\frac{e^{-(t+h)x}-e^{-tx}}{hx}dx$
($t>0$ is a constant.)
I know the limit in the right side is $-e^{-tx}$ if that matters.
Now, by Heine, it's sufficient to prove this result with a sequence $h_n\rightarrow 0$ instead  of a continuous variable. So let's define $f_n(x)=\frac{e^{-(t+h_n)x}-e^{-tx}}{h_nx}$ and now we need to show that the conditions of DCT hold, and all that's left is finding a dominating function.
That's where I failed. Any idea?
 A: Fix $h >0$. Note that 
$$  \frac{e^{-(t+h)x}-e^{-tx}}{hx} = e^{-tx} \left( \frac{e^{-hx}-1}{hx} \right) 
\implies  \left\vert \frac{e^{-(t+h)x}-e^{-tx}}{hx} \right\vert = e^{-tx}
\left\vert \frac{e^{-hx}-1}{hx} \right\vert $$
Now, define $f(x) = e^{-x}$. Then $$\frac{e^{-hx}-1}{hx} = \frac{f(hx) - f(0)}{hx}$$
and by Mean Value Theorem
$$
\frac{f(hx) - f(0)}{hx} = f'(\xi_x) = -e^{-x} \Big|_{x = \xi_x} = -e^{-\xi_x} 
$$
for some $\xi_x \in (0,hx)$.
Now define $g(y)  = -e^{-y}$ for $y>0$. It is clear that $|g(y)| < 1$ for all $y >0$.
So for all $x$ we have
$$
\left\vert \frac{ f(hx) - f(0)}{hx} \right\vert = |-e^{-\xi_x}| \le 1
$$
which means
$$
\left\vert \frac{e^{-(t+h)x}-e^{-tx}}{hx} \right\vert =  e^{-tx}
\left\vert \frac{e^{-hx}-1}{hx} \right\vert 
\le e^{-tx} 
$$
and the RHS is integrable on $[0,\infty)$.
EDIT: If $h<0$, then write 
$$
 \frac{e^{-(t+h)x}-e^{-tx}}{hx} = e^{-(t+h)x} \left( \dfrac{ 1 - e^{hx}}{hx} \right)$$
and
$$  \dfrac{ 1 - e^{hx}}{-hx} = \dfrac{ g(0)- g(hx)}{-hx}
$$
where $g(x)  = e^{x}$. Then, by MVT we get
$$\dfrac{ g(0)- g(hx)}{-hx} = g'(\xi_x) = e^{\xi_x}
$$
for some $\xi_x$ in $(hx,0)$. Then, $e^{\xi_x} \le 1$ because of this interval.
We can further restrict ourselves to $-\frac{t}{2} < h < 0$, since we're taking a limit and only care about small $h$. Finally
$$
\left\vert e^{-(t+h)x} \left( \dfrac{ 1 - e^{hx}}{hx} \right)
\right\vert \le e^{-(t+h)x} \le e^{-tx/2}
$$
and this serves as a dominating function.
A: A very efficient way forward circumvents use of the Dominated Convergence Theorem and instead exploits Frullani's Theorem.  Proceeding, we obtain 
$$\frac1h \int_0^\infty \frac{e^{-(t+h)x}-e^{-tx}}{hx} \,dx=\frac1h \log\left( \frac{t}{t+h}\right)$$
The limit as $h\to 0$ is straightforward to evaluate with the answer $-\frac1t$.  And we are done!
If one wishes to proceed through use of the Dominated Convergence Theorem, then the following primer will facilitate analysis.

PRIMER: 
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities 
$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}} \tag 1$$
for $x<1$.  


Herein, we use the inequality on the left-hand side of $(1)$ to establish a coveted dominating function.  To that end, we proceed.
Note that we can write the integrand $\frac{e^{-(t+h)x}-e^{-tx}}{hx}$ as
$$\begin{align}
\frac{e^{-(t+h)x}-e^{-tx}}{hx}&=e^{-tx}\left(\frac{e^{-hx}-1}{hx}\right) \tag{2a}\\\\
&=e^{-(t+h)x}\left( \frac{1-e^{hx}}{hx} \right) \tag{2b}
\end{align}$$
Note that the term in parentheses on the right-hand side of $(2a)$ is non-positive for all $x\in [0,\infty$ and all $h$.  Applying the left-hand side inequality in $(1)$ to $(2a)$, we find immediately that for $h>0$
$$\left|\frac{e^{-hx}-1}{hx}\right|\le 1$$
for $x\ge 0$, $h>0$.
For $h<0$, we take $|h|<t/2$.  Then, applying the left-hand side inequality in $(1)$ to $(2b)$, we see analogously that
$$\left| \frac{1-e^{hx}}{hx} \right|\le 1$$
for $x\ge 0$, $h<0$.
Hence, we have 
$$\left|\frac{e^{-(t+h)x}-e^{-tx}}{hx}\right|\le e^{-tx/2}$$
for all $x\ge 0$ and all $|h|\le t/2$.
Inasmuch as $\int_0^\infty e^{-tx/2}\,dx=\frac2t <\infty$, the Dominated Convergence Theorem guarantees that
$$\begin{align}
\lim_{h\to 0}\int_0^\infty \frac{e^{-(t+h)x}-e^{-tx}}{hx}\,dx&=\int_0^\infty \lim_{h\to 0}\left(\frac{e^{-(t+h)x}-e^{-tx}}{hx}\right)\,dx\\\\
&=-\int_0^\infty e^{-tx}\,dx\\\\
&=-\frac1t
\end{align}$$
And we are done!  
