Finding the points at which $y=e^x-1$ and $y=x$ meet without using derivatives.

Question

I was trying to solve this limit without using derivatives:

$$\lim_{x\rightarrow 0}\frac{e^x-1}{x}$$

I was thinking of using graphs and going for the solution, but I wasn't able to prove that $e^x-1$ and $x$ meet only once, at $x=0$ rigorously without finding slopes using derivatives. Is there a way around this that I'm not able to see?

(Side note: I was trying to prove that $\frac{d}{dx}e^x$ is $e^x$ without both Taylor expansions and taking derivative of both the numerator and denominator (for L'Hôpital))

Answer 1

As, $e$ can be defined with limits, I will use $e$ without any complicated proofs.

$e^x - 1= x$

$\displaystyle \rightarrow \frac{e^x-1}{x-0} = 1$

Let $f(x) = e^x$

$\displaystyle \rightarrow \frac{f(x)-f(0)}{x-0} = 1$

LHS implies the gradient of the line connecting $(0, f(0))$ and $(x, f(x))$.

As $e^x$ is a convex function, gradient of the line is strictly monotonically increasing,

which means that the $x$ satisfying the equations should be unique and it's $0$.

Answer 2

Here is an alternative method: \begin{equation} \lim_{x \to 0}\ \dfrac{e^x-1}{x}=\lim_{t \to 0}\ \dfrac{t}{\log(1+t)}= \lim_{s \to \infty}\ \dfrac{1}{s\ \log\left(1+\dfrac{1}{s}\right)}=\lim_{s \to \infty}\ \dfrac{1}{ \log\left(1+\dfrac{1}{s}\right)^s} \end{equation} where $t=e^x-1$ and $s=\dfrac{1}{t}$. Now using the fact: \begin{equation} \lim_{s \to \infty} \left(1+\dfrac{1}{s}\right)^s=e \end{equation} we have finally \begin{equation} \lim_{x \to 0}\ \dfrac{e^x-1}{x}=\dfrac{1}{\log(e)}=1 \end{equation}