how do you differentiate $\ln(x)$ using the difference quotient. So the limit is as $h$ approaches $0$ of $\displaystyle \frac{\ln(x+h)-\ln(x)}{h}$, which simplifies to $\displaystyle\frac{\ln\left(\frac{x+h}{x}\right)}{h}$, which simplifies to $\displaystyle\frac{\ln\left(1+ \frac hx\right)}{h}$.
I got stuck here.  , how should I continue?
 A: For finite non-zero $x,$ $$\lim_{h\to0}\dfrac{\ln\left(1+\dfrac hx\right)}h=\dfrac1x\cdot\lim_{h\to0}\dfrac{\ln\left(1+\dfrac hx\right)}{\dfrac hx}=?$$
A: In THIS ANSWER, I used only the limit definition of the exponential function and Bernoulli's Inequality to show that the logarithm function satisfies the inequalities 
$$\frac{x-1}{x}\ge \log(x)\le x-1\tag1$$
Hence, using $\log(x+h)-\log(x)=\log\left(1+\frac hx\right)$, we from $(1)$ we have 
$$\frac{\frac1x}{1+\frac hx}\le \frac{\log\left(1+\frac hx\right)}{h}\le \frac1x\tag2$$
Applying the squeeze theorem to $(2)$ we find the coveted limit
$$\lim_{h\to 0}\frac{\log(x+h)-\log(x)}{h}=\frac1x$$
A: If the definition of $\ln x$ is 
$$\ln x = \int_1^x \frac{1}{t} \; dt$$
Then you can write your limit as
$$\lim_{h \to 0 } \frac{1}{h} \int_x^{x+h} \frac{1}{t} \; dt.$$
Then note that for very small $h$, the area represented by the integral is nearly a rectangle of width $h$ and height $1/x$, so the limit equals
$$\lim_{h \to 0} \frac{1}{h}(h/x) = \frac{1}{x}.$$
A: $$\lim_{h\to 0} \frac{\ln(1+\frac hx)}{h}=\frac1x\lim_{\frac hx \to 0}\frac{\ln(1+\frac hx)}{\frac hx}=\frac1x$$
A: Multiply and divide the denominator by $x$. Clearly we won't differentiate at $x=0$.
So,
$$L = \lim_{h\to0}\frac{\ln(1+\frac{h}{x})}{x\cdot\frac{h}{x}} = \frac{1}{x}\lim_{\frac{h}{x}\to0}\frac{\ln(1+\frac{h}{x})}{\frac{h}{x}} = \frac{1}{x}\cdot1 = \frac{1}{x}$$
A: Option:
$x >0$
$\log x=\displaystyle{\int_{1}^{x}} (1/t )dt.$
MVT for integration.
$(1/h)\dfrac{\log (x+h)-\log x}{h}=$
$(1/h)\displaystyle{\int_{x}^{x+h}}(1/t)dt=$
$(1/h)(1/s)\displaystyle{\int_{x}^{x+h}}1 dt=$
$(1/h)(1/s)h$; where $s \in [x,x+h].$
Take the limit $h \rightarrow 0.$
