I want to prove that $$\lim_{x \to -1^{-}}\log(1+\frac{1}{x})- \frac{1}{1+x} = +\infty.$$

I know that (but I can't prove it also) $$\lim_{x\to -1^{-}}\log(1+\frac{1}{x})=-\infty.$$ and $$\lim_{x\to -1^{-}}\frac{-1}{1+x}=+\infty.$$ which means that I have an indeterminate form $-\infty+\infty$ but I don't know how to get over this IF.

# Proof $\lim_{x \to -1^{-}}\log(1+\frac{1}{x})- \frac{1}{1+x}= +\infty$ without L’hôpital’s rule

calculuslimitslogarithms

## Best Answer

Let $$ u=-\frac1{x+1} $$ and then $x=-\frac{u+1}{u}$. So $$ \log(1+\frac{1}{x})- \frac{1}{1+x} = u-\log(u+1) $$ and hence if $x\to-1^-$, then $u\to\infty$. For $u>6$, $$ e^u=1+u+\frac{u^2}{2}+\frac{u^3}{3!}+\cdots>1+u+\frac{u^2}{2}+\frac{u^3}{3!}>u^2+2u+1=(u+1)^2. $$ So $$ u-\log(u+1)>\log(u+1) $$ and hence $$\lim_{x \to -1^{-}}\log(1+\frac{1}{x})- \frac{1}{1+x} = \lim_{u\to\infty}(u-\ln(u+1))=\infty.$$