How to compute $\lim_{x\to \infty} \frac{\ln(1+x)}{\ln x} $?

Question

How to compute $\lim_{x\to \infty} \frac{\ln(1+x)}{\ln x} $? I managed to find it is $1$ by using L'Hôpital's, but I am sure there is also an elementary solution.

Answer 1

Hint: $$ \frac{\ln(x+1)}{\ln x}=1+\frac{\ln (x+1)-\ln x}{\ln x}=1+\frac{\ln (1+\frac1{x})}{\ln x}. $$

Answer 2

$\forall x>0, \log(x+1)=\log(x)+\log(1+\frac{1}{x})$ thus:$$ \frac{\log (x+1)}{\log(x)}=1+\frac{\log(1+\frac{1}{x})}{\log x}\rightarrow 1$$

Answer 3

$x >1;$

$1= \dfrac{\log (1+x)}{\log (1+x)} \lt \dfrac{\log (1+x)}{\log x} \lt $

$\dfrac{\log (2x)}{\log x} =$ $\dfrac{\log 2}{\log x }+ 1.$

Take the limit.

Answer 4

Here is a way with substitution $x = e^y$ and squeezing using the inequality

• $1+e^y < e^{1+y}$ for $y>0$:

\begin{eqnarray*} 1 & = & \frac{\ln (1+x)}{\ln(1+x)} \\ & < & \frac{\ln (1+x)}{\ln x} \\ & \stackrel{x=e^y}{=} & \frac{\ln (1+e^y)}{y} \\ & \stackrel{1+e^y < e^{1+y}}{<} & \frac{\ln (e^{1+y})}{y} \\ & = & \frac{1+y}{y} \\ & \stackrel{y \to +\infty}{\longrightarrow} 1 \end{eqnarray*}