Solve $\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x}$ without using L'Hopital or Taylor Series 
Solve $$\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x}$$ without using L'Hopital or Taylor Series

I tried:
\begin{align}\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x} &=
\frac{\ln(x+1)}{x}-\frac{\ln(x)}{x} \\
&=\lim \frac{\ln(x+1)}{x}-\lim\frac{\ln(x)}{x} \\
&=\lim \frac{\ln(x+1)}{x}-0 \\
&=\lim \frac{\ln(x+1)}{x} \\
&=\lim_{x \rightarrow 0} \frac{\ln\left(\frac{1}{x}+1\right)}{\frac{1}{x}} \\
&=\lim_{x \rightarrow 0} x\ln\left(\frac{1}{x}+1\right) \\
&=\lim_{x \rightarrow 0} \ln\left(\left(1+\frac{1}{x}\right)^x\right) \\
&= ???\end{align}
How do I solve this? What do I do next?
 A: The denominator clearly $\to\infty$
Now for the numerator using $\log a/b=\log a-\log b$ for all the logarithm remain defined,
$$\lim_{x\to\infty}(\ln(1+x)-\ln x)=\lim_{x\to\infty}\ln\left(1+\dfrac1x\right)=?$$
Alternatively,
Set $1/x=h$  to get $$\lim_{h\to0}h\cdot\ln(1+h)$$
A: Recall that 
$$ \ln x := \int_1^x \frac{1}{t} ~dt.$$
Thus 
\begin{align}
 \lim_{x \to \infty} \frac{\ln (x+1) - \ln x}{x} &= \lim_{x \to \infty} \frac{1}{x} \left[ \int_1^{x+1} \frac{1}{t} ~dt - \int_1^x \frac{1}{t} ~dt \right] \\
 &= \lim_{x \to \infty} \frac{1}{x}  \int_x^{x+1} \frac{1}{t} ~dt
\end{align}
A: $$\lim_{x\to +\infty}\frac{\log(x+1)-\log(x)}{x}=\lim_{y\to +\infty}\frac{\log(1+e^y)-y}{e^y}=\lim_{y\to +\infty}\frac{\log(1+e^{-y})}{e^y}=0$$
by squeezing: for any $x>0$ we have $\log(1+x)\leq x$ and $e^x\geq x+1$.
A: Let $f(x)=\ln x$. Now use Lagrange's MVT in $[x,x+1]$ to get
$$ f(1+x)-f(x)=f'(\xi), \xi\in(x,1+x) $$
or
$$ \ln(1+x)-\ln x=\frac1{\xi}.$$
So
$$ \lim_{x\to\infty}\frac1x\ln\frac{1+x}x=0.$$
A: For $x>1,$
$$0 <\frac{\ln(x+1) - \ln x}{x} < \frac{\ln(x+1)}{x} < \frac{\ln (2x)}{x} = \frac{\ln 2 + \ln x}{x}.$$
Because $(\ln x)/x \to 0$ (a result you appear to know), the limit is $0$ by the squeeze theorem.
