Q:

How to deduce the limit relation $\lim_{x\to0} \frac{e^{cx}-1}{x}=c$

Let $f(x) = e^{cx}$ where $c$ is constant. Show that $f'(0)=c$ and use this to deduce the limit relation $$\lim_{x\to0} \frac{e^{cx}-1}{x}=c$$
Proving $f'(0)=c$ is easy but I'm not sure how the limit is proven. 

A:

Hint: If you were to apply the definition of the derivative to the function $f$ at the point $x = 0$, which limit would you have to calculate?

A:

I thought it might be instructive to present an approach that does not rely on differential calculus, but rather relies only on the inequalities for $x<1$
$$1+x\le e^x\le \frac{1}{1-x}$$
which I showed in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality.
Then, we have
$$c\le \frac{e^{cx}-1}{x}\le \frac{c}{1-cx}$$
whereupon applying the squeeze theorem yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{e^{cx}-1}{x}=c}$$

A:

It is the definition of the derivative : $$f'(0)=\lim_{x\to0} \frac{f(x)-f(0)}{x-0}$$
Can you go from there ?

