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

Question

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.

Answer 1

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?

Answer 2

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

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{e^{cx}-1}{x}=c}$$

Answer 3

It is the definition of the derivative : $$f'(0)=\lim_{x\to0} \frac{f(x)-f(0)}{x-0}$$

Can you go from there ?

Answer 4

Hint. One may recall that, for any differentiable function $f$ near $a$, one has

$$ \frac{f(x)-f(a)}{x-a}\to f'(a) $$

as $x \to a$.

Answer 5

For $0 < x < 1$, $1+x < e^x < \dfrac1{1-x}$.

Therefore, if $0 < x < \dfrac1{c}$, $1+cx < e^{cx} < \dfrac1{1-cx} $ or $cx < e^{cx}-1 < \dfrac1{1-cx}-1 =\dfrac{cx}{1-cx} $ or

$\begin{array}\\ c &< \dfrac{e^{cx}-1}{x}\\ &<\dfrac{c}{1-cx}\\ &=c+\dfrac{c^2x}{1-cx}\\ &<c+2c^2x \qquad\text{for } cx < \frac12 \text{ or } x < \frac1{2c}\\ \end{array} $

Letting $x \to 0$, $\lim_{x \to 0^+} \dfrac{e^{cx}-1}{x} = c $.

Answer 6

Look at the series expansion:

$$e^{cx} = 1 + \sum_{n=1}^\infty \frac{(cx)^{n}}{n!} $$

Subtract one and dividing by x leaves:

$$\sum_{n=1}^\infty \frac{(cx)^{n-1}}{n!} = c + \sum_{n=2}^\infty \frac{(cx)^{n-1}}{n!}$$

All terms for n>=2 contain some nonzero power of x and approach zero as x approaches zero.