How to effectively rewrite $\lim_{x \to 0} \frac{e^{x}-1}{x}$ to properly assess its value

Question

I am tasked with evaluating the following limit:

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

There are two things I have tried. The first, substituting $x$ for $\ln t$ gave me

$$\lim_{x \to o} \frac{t-1}{\ln t}$$

But I did not know what to do from there. The other attempt was using the numerator's conjugate like so:

$$\lim_{x \to 0} \frac{t^{2}-1}{\left(t+1\right)\ln t}$$

But again, I did not know where to go from there. What do you suggest I try?

Cheers

Answer 1

Use L'Hôpital's rule

$\begin{align}&\lim_{x \to 0} \frac{(e^{x}-1)}{x}\quad(\frac{0}{0}-\text{form})\\&= \lim_{x \to 0} \frac{e^{x}}{1}\\&=e^0\\&=1 \end{align}$

Answer 2

Let $x=\log (t+1)$, so when $x\to 0$ than $t\to 0$, and the limit converts into: $$\lim_{x\to 0}\frac{e^x-1}{x}=\lim_{t\to 0}\frac{t}{\log (t+1)}=\lim_{t\to 0}\frac{1}{\frac{\log(t+1)}{t}}=\\ \lim_{t\to 0}\frac{1}{\frac{1}{t} \log(t+1)}=\lim_{t\to 0}\frac{1}{\log (1+t)^{\frac{1}{t}}}=\frac{1}{\log e}=1$$

Since $$e:=\lim_{x\to\infty}(1+\frac{1}{x})^x=\lim_{t\to 0}(1+t)^{\frac{1}{t}}$$

Answer 3

For these kinds of limits, whenever the quotient rule yields "0/0", you can either l'Hopital's Rule or Taylor Series.

Using l'Hopital's Rule, we have

$$ \lim_{x\to0}\frac{e^x-1}{x}=\lim_{x\to0}\frac{(e^x-1)'}{(x)'}=\lim_{x\to0}\frac{e^x}{1}=1. $$

Since the $n$-th Taylor polynomial of $e^x$ about $x=0$ is given by $$ T_n(x)=1+x+\frac{x%2}{2!}+\cdots+\frac{x^n}{n!}. $$ we have, for $n\ge 2$: $$ \lim_{x\to0}\frac{e^x-1}{x}=\lim_{x\to0}\frac{T_n(x)-1}{x}=\lim_{x\to0}\left[1+\frac{x}{2!}+\cdots+\frac{x^{n-1}}{n!}\right]=1. $$

Alternatively, one can also notice that if $f(x)=e^x$, then the limit $$ \lim_{x\to0}\frac{e^x-1}{x}=\lim_{x\to0}\frac{f(x)-f(0)}{x} $$ represents the derivative of $f$ at $x=0$. Since $f'(x)=e^x$, we get $$ \lim_{x\to0}\frac{e^x-1}{x}=\lim_{x\to0}\frac{f(x)-f(0)}{x}=f'(0)=e^0=1. $$