Prove that $\displaystyle \lim_{x\rightarrow\infty}\frac {\ln (x+1)}{\ln x}=1$.
Prove that $\displaystyle \lim_{x\rightarrow\infty}\frac {\ln (x+1)}{\ln x}=1$.
$\displaystyle \lim_{x\rightarrow\infty}\frac {\ln (x+1)}{\ln x}= \lim_{x\rightarrow\infty}\frac {\frac{1}{1+x}}{\frac{1}{x}}= \lim_{x\rightarrow\infty}\frac {x}{x+1}=1$