How would I use the definition of derivative to prove $$\lim_{x\to 0} \frac{\ln(1+x)}{x} = 1$$
I got to $$\frac{\frac{\ln(1+x+h)}{(x+h)} - \frac{\ln(1+x)}{x}}{h}$$ but have no idea where to go from here.
On another site I found someones answer where they stated the following: $$ \lim_{x\to 0} \frac{\ln(1+x) - \ln(1+0)}{x-0} = [\ln(1+x)]'\rvert_x = 0 $$ but I am unsure why the $x$ in the $x-0$ is removed. Can someone please explain?
$$\lim_{x\to0}\frac{\ln(x+1)}{x}=\lim_{x\to0}\frac{\ln(x+1)-\ln 1}{x}=(\ln t)'|_{t=1}=1$$
$$ \begin{align} \lim_{x\to 0} \frac{\ln(1+x)}{x} &= \\ &= \lim_{x\to 0} \frac{1}{x}\ln(1+x) \quad\quad\quad\text{(Using log rules:)}\\ &= \lim_{x\to 0} \ln\left((1+x)^{\frac{1}{x}}\right) \\ &= \lim_{x\to 0} \ln\left((1+\frac{1}{\frac{1}{x}})^{\frac{1}{x}}\right) \\ &= \begin{bmatrix}b = \frac{1}{x} \\ x \to 0 \implies b \to \infty \end{bmatrix} \\ &= \lim_{b\to \infty} \ln\left((1+\frac{1}{b})^{b}\right) \\ &= \ln\left(\lim_{b\to \infty}(1+\frac{1}{b})^{b}\right) \\ &= \ln(e) \\ &= 1 \end{align} $$
from the definition: $$f'(x_0)=\lim_{x\rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}$$
so if I say that $f(x)=ln(1+x)$ we know that $f'(x)=\frac{1}{x+1}$ $$ 1=f'(0) = \lim_{x\rightarrow0}\frac{f(x)-f(0)}{x-0} = \lim_{x\rightarrow0}\frac{f(x)-f(0)}{x} = \lim_{x\rightarrow0}\frac{f(x)}{x} = \lim_{x\rightarrow0}\frac{ln(x+1)}{x}$$
remember that $f(0)=ln(1)=0$
