<s>[INST] 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?
 [/INST] 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}
</s>