Deriving results from a known limit Assuming that we know $$\lim_{x\rightarrow 0}\frac{e^x-1}{x}=1$$ is it possible to derive that $$\lim_{x\rightarrow 0}\frac{e^{-ax}-e^{-bx}}{x}=b-a$$
$$\lim_{x\rightarrow 0}\frac{\tanh{ax}}{x}=\lim_{x\rightarrow 0}\frac{\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}}{x}=a$$
without using L'Hôpital's rule? 
For the first one I tried separating it into 
$$\lim_{x\rightarrow 0}\left(\frac{e^{-ax}-1}{x}-\frac{e^{-bx}-1}{x}\right)$$
But I don't know how to continue.
 A: $$\lim_{x\rightarrow 0}\frac{e^{-ax}-e^{-bx}}{x}= \lim_{x\rightarrow 0}\left(-a\frac{e^{-ax}-1}{-ax}-(-b)\frac{e^{-bx}-1}{-bx}\right) = b-a$$
$$ \lim_{x\rightarrow 0}\frac{\tanh{ax}}{x}=\lim_{x\rightarrow 0}\frac{\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}}{x}= \lim_{x\rightarrow 0}\frac{e^{ax}-e^{-ax}}{x} \cdot \frac{1}{e^{ax}+e^{-ax}}=  2a \cdot \frac{1}{2} = a$$
A: As we are given that
$\lim_{x\rightarrow 0}\dfrac{e^x-1}{x} = 1, \tag{1}$
setting
$x = \alpha y, \tag{2}$
$\alpha$ constant, we have
$\dfrac{1}{\alpha} \lim_{y \rightarrow 0}\dfrac{e^{\alpha y}-1}{y} = \lim_{y \rightarrow 0}\dfrac{e^{\alpha y}-1}{\alpha y} = \lim_{\alpha y \rightarrow 0}\dfrac{e^{\alpha y}-1}{\alpha y} = 1, \tag{3}$
so that
$\lim_{y \rightarrow 0}\dfrac{e^{\alpha y}-1}{y} = \alpha. \tag{4}$
Applying (3) to
$\lim_{x\rightarrow 0}\dfrac{e^{-ax}-e^{-bx}}{x}, \tag{5}$
we have
$\lim_{x\rightarrow 0}\dfrac{e^{-ax}-e^{-bx}}{x} = \lim_{x\rightarrow 0} e^{-bx} \dfrac{e^{(b - a)x}-1}{x} = (\lim_{x\rightarrow 0} e^{-bx})(\lim_{x\rightarrow 0}\dfrac{e^{(b - a)x}-1}{x})$
$= 1(b - a) = b - a, \tag{6}$
establishing the first requisite limit.  The second follows easily from the first:
$\lim_{x\rightarrow 0}\dfrac{\tanh{ax}}{x}=\lim_{x\rightarrow 0}\dfrac{\dfrac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}}{x} = (\lim_{x\rightarrow 0} \dfrac{e^{ax}-e^{-ax}}{x})(\lim_{x\rightarrow 0} \dfrac{1}{e^{ax}+e^{-ax}})$
$= (a - (-a))\dfrac{1}{2} = a. \tag{7}$
Hope this helps!  Solstice-Time Greetings,
and as ever,
Fiat Lux!!!
