Determine the following limit as x approaches 0: $\frac{\ln(1+x)}x$ $$\lim_{x\to 0} \frac{\ln(1+x)}x$$
The process I  want to take to solving this is by using the definition of the limit, but I am getting confused. ( without l'hopitals rule) 
$$\lim_{h \to 0} \frac{f(x+h) - f(x)}h$$
$$\lim_{h \to 0} \frac{\frac{\ln (1+x+h)}{x+h} - \frac{\ln(1+x)}x}h$$
$$\lim_{h \to 0} \frac{x\ln(1+x+h) - (x+h)\ln (1+x)}{hx(x+h))}$$
At this point I get confused because I know the answer is $1$, but I am not getting this answer through simplification of my formula.
 A: 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 $\log(x)$)
A: 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 $\ln(e) = 1$
A: 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

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

