calculus - 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.

Answer

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 $\log(x)$)