How to compute $\lim\limits_{x\to 0}\dfrac{e^x-1}{x}$? How does one compute $\lim\limits_{x\to 0}\dfrac{e^x-1}{x}$ without using l'Hôpital's rule or knowledge about the derivative of $e^x$ ? $e^x$ denotes the exponential function with base $e$ (Euler's constant).
 A: Using the series definition of the exponential function one obtains:
$\dfrac{e^x - 1}{x} = \dfrac{\displaystyle\sum\limits_{k = 0}^{\infty}\dfrac{x^k}{k!}-1}{x} = \dfrac{\displaystyle\sum\limits_{k = 1}^{\infty}\dfrac{x^k}{k!}}{x} =
{\displaystyle\sum\limits_{k = 2}^{\infty}\dfrac{x^{k-1}}{k!}} =
 1+{\displaystyle\sum\limits_{k = 2}^{\infty}\dfrac{x^{k-1}}{k!}}  $.
Hence,
$\lim\limits_{x\to 0}\dfrac{e^x - 1}{x}=\lim\limits_{x\to 0} \left(1+{\displaystyle\sum\limits_{k = 2}^{\infty}\dfrac{x^{k-1}}{k!}}\right) = 1 + \lim\limits_{x\to 0} {\displaystyle\sum\limits_{k = 2}^{\infty}\dfrac{x^{k-1}}{k!}} \stackrel{(1)}{=} 1 + 0 = 1.$
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(1)  $\lim\limits_{x\to 0} {\displaystyle\sum\limits_{k = 2}^{\infty}\dfrac{x^{k-1}}{k!}} = \lim\limits_{x\to 0} {\displaystyle\sum\limits_{k = 0}^{\infty}\dfrac{x^{k+1}}{(k+2)!}} = \lim\limits_{x\to 0} x{\displaystyle\sum\limits_{k = 0}^{\infty}\dfrac{x^{k}}{(k+2)!}} = 0 \cdot \dfrac{1}{2} = 0.$
$\phantom{\\}$
Alternative approach:
We make the substitution $u = e^x-1$ to obtain
$\lim\limits_{x\to 0}\dfrac{e^x - 1}{x}=\lim\limits_{u\to 0} \dfrac{u}{\ln(u+1)} = \lim\limits_{u\to 0} \dfrac{1}{\dfrac{1}{u}\ln(u+1)} = \lim\limits_{u\to 0}\dfrac{1}{\ln((u+1)^{1/u})}$
Now when  substituting $t = 1/u$ it becomes obvious that
$\lim\limits_{u\to 0}\dfrac{1}{\ln((u+1)^{1/u})} = \lim\limits_{t\to \infty}\dfrac{1}{\ln((1/t+1)^{t})} \stackrel{(2)}{=} \left(\ln\left(\lim\limits_{t\to \infty} \left(1 + \dfrac{1}{t}\right)^{\large t}\right)\right)^{\large-1} = (\ln(e))^{-1} = 1.$
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(2) We use the fact that $\ln(x)$ is continuous at $x=e$.
A: The evaluation of the limit will heavily depend on how you've defined the exponential function $e^{(\cdot)}:\mathbb{R}\to\mathbb{R}$ (I'll use the $\exp$ notation here). If it's defined as the unique solution of the initial value problem $f'=f,f(0)=1$, then the limit immediately follows from the fact that $(\exp)'(0)=\exp(0)=1$ and the limit definition of the derivative.
\begin{align*}
1 &= (\exp)'(0)\\
&= \lim\limits_{h\to 0}\frac{\exp(0+h)-\exp(0)}{h}\\
&= \lim\limits_{h\to 0}\frac{\exp(h)-1}{h}
\end{align*}
If $\exp(x)$ was defined as the sum of the power series $\sum_{n=0}^{\infty}\frac{x^n}{n!}$, then the limit can be established by making use of the continuity of power series:
\begin{align*}
\lim\limits_{h\to 0}\frac{\exp(h)-1}{h} &= \frac{\left(1+h+\frac{h^2}{2!}\cdots\right)-1}{h}\\
&= \lim\limits_{h\to 0}\frac{h+\frac{h^2}{2!}+\frac{h^3}{3!}\cdots}{h}\\
&=  \lim\limits_{h\to 0}\left(1+\frac{h}{2!}+\frac{h^2}{3!}\cdots\right)\\
&= 1
\end{align*}
Here are a few more characterizations:

*

*$\exp(x):=\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n$

*$\exp$ is the inverse function of $\ln$, where $\ln$ is defined in a way that avoids circularity, e.g. $\ln(x)=\int_{1}^{x}\frac{1}{t}dt$.

*$\exp$ is the unique function $f$ that is continuous at some point, satisfies $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb{R}$, and satisfies $f(1)=e$, where $e$ is defined in a way that avoids circularity, e.g. $e=\lim_{n\to\infty}(1+1/n)^n$
I would LOVE prove that $\lim_{h\to 0}[\exp(h)-1]/h=1$ from one of these characterizations. Unfortunately, my current level of mathematical knowledge prohibits me from doing so. Also, this response was more to draw your attention to the point that different definitions lead to different proofs. Definitions are important!
A: Define $\,a_n:=n(b^{1/n}-1).\,$ Verify that
$$a_n=\frac{n(b-1)}{(b-1)/(b^{1/n}-1)} =
\frac{n(b-1)}{1+b^{1/n}+b^{2/n}+\cdots+b^{(n-1)/n}}. $$
Notice that
$$ \lim_{n->\infty} \frac{1+b^{1/n}+b^{2/n}+\cdots+b^{(n-1)/n}}n
 = \int_0^1 b^x dx = \frac{b-1}{\log b} $$
using Riemann sums and that $\,\int b^x dx =\dfrac{ b^x}{\log b}+C.\,$ Thus,
$$ \lim_{x\to0} \frac{b^x-1}x = \lim_{n\to\infty} a_n = \log b. $$
A: Let $x\in\mathbb{R}$ such that $0<|x|<1$: By Bernoulli's inequality we have
\begin{align}
e^{+x}&\geq 1+x\tag{1}\\
e^{-x}&\geq 1-x\tag{2}
\end{align}
Inequality $(2)$ is equivalent to
$$e^{x}\leq \frac{1}{1-x}\tag{3}$$
From $(1)$ and $(3)$ we obtain
$$1\leq \frac{e^x-1}{x}\leq \frac{1}{1-x}$$
Squeeze theorem implies then
$$\lim_{x\to 0} \frac{e^x-1}{x}=1$$
A: We have :
$$\lim_{x\to 0} \frac{e^x-1}{x}=\lim_{x\to 0}\frac{e^x-e^0}{x-0}$$
If we considered the following function : $f(x)=e^x$ then :
$$f'(0)=\lim_{x\to 0}\frac{e^x-e^0}{x-0}$$
And the derivative of the exponential function is itself, so $f'(0)=e^0=1$, and your limit is :
$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$
