Determine the following limit as x approaches 0: $\frac{\ln(1+x)}x$

You are talking about L'Hôpital's rule, so I assume you already know how to differentiate the logarithm. Now note, that

$$\frac{\log(x+1)}x = \frac{\log(x+1)-\log(1)}{(x+1)-1}$$

Thus

$$\lim_{x\to0}\frac{\log(x+1)}x = \lim_{x\to0}\frac{\log(x+1)-\log(1)}{(x+1)-1}=\left(\log(x)\right)^\prime_{x=1}=\left.\frac{1}x\right|_{x=1}=1$$

(This is not by using L'Hôpital's rule but only by using the definition of derivative and knowing the derivative of $\log(x)$)

Introduce a new variable $u = 1/x$. Then you limit becomes

$$\lim_{u\to\infty} \frac{\ln(1 + 1/u)}{1/u}$$

mulitply numerator and denominator by $u$, you get

$$\lim_{u\to\infty} u [ \ln(1 + 1/u) ] $$

move $u$ into the $\log$, getting

$$\lim_{u\to\infty} [ \ln(1 + 1/u)^u ]$$

then, since $\ln$ is continuous

$$\ln \left( \lim_{u\to\infty} (1 + 1/u)^u \right)$$

limit inside is equal to $e$ and $\ln(e) = 1$

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag 1$$

for $x>0$.

Hence, we have from $(1)$

$$\frac{1}{1+x}\le \frac{\log(1+x)}{x}\le 1$$

whereupon application of the squeeze theorem yields the coveted result

$$\lim_{x\to 0}\frac{\log(1+x)}{x}=1$$