(Please think of a pun for LIMIT yourself)

Usually L'Hopital's Rule is the way i would go, but i decided to take a different route.

The Taylor Seriers expansion of $\ln(x+1)$ at $x=0$ is $\displaystyle\sum_{n=0}^{\infty}(-1)^{n+1}\frac{x^n}{n} = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$

We can plug this into our limit and we can see that: $\begin{aligned} \lim\limits_{x\to 0}\frac{\ln(x+1)}{x} &\Longrightarrow \lim\limits_{x\to 0}\frac{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots}{x} \\ &= \lim\limits_{x\to 0}1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\dots \\ &= 1-\frac{0}{2}+\frac{0^2}{3}-\frac{0^3}{4}+\dots = \boxed{1} \end{aligned}$

$\text{We first substitute } x=0$

$\large{\lim \limits_{x\rightarrow 0} \dfrac{\ln{(x+1)}}{x} =\dfrac{0}{0}}$

$\text{We get a } \dfrac{0}{0} \text{ situation, so we can apply L'Hôpital's rule.}$

$\text{We differentiate both numerator and denominator to get:}$

$\large{\lim \limits_{x\rightarrow 0} \dfrac{\ln{(x+1)}}{x} = \dfrac{\frac{d}{dx} (\ln{(x+1)})}{\frac{d}{dx} (x)}=\dfrac{1}{x+1}}$

$\text{Now we can substitute } x=0 \text{ and get the answer.}$

$\large{\lim \limits_{x\rightarrow 0} \dfrac{\ln{(x+1)}}{x} = \dfrac{1}{0+1}=\boxed{1}}$

So as you can see, putting x=0 does not help, you become sick, so you need to go to the Hospital. Idk what pun to use.

Let $f(x)=\ln x$

Then $f'(x) =\dfrac 1x$

So, $f'(1)=\dfrac 11=1=$

$\displaystyle \lim_{x\to 0} \dfrac {\ln (1+x)-\ln 1}{x}$

$=\displaystyle \lim_{x\to 0} \dfrac {\ln (1+x)}{x}$