Livia Cardenas

2022-07-22

L'Hôpital's as x tends to infinity
I'm searching for the explanation to the limit of:
$\underset{x\to \mathrm{\infty }}{lim}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{ln}\frac{x+1}{x-1}.$
I know the answer is 2, but I can't seem to get there. The problem is in my textbook under a section with l'Hôpital.

Kitamiliseakekw

Expert

Hint:
$\mathrm{ln}\frac{x+1}{x-1}=\mathrm{ln}\left(x+1\right)-\mathrm{ln}\left(x-1\right)$
so that
$x\cdot \mathrm{ln}\frac{x+1}{x-1}$
can be rewritten as
$\frac{\mathrm{ln}\left(x+1\right)-\mathrm{ln}\left(x-1\right)}{{x}^{-1}}.$
Now,
$\underset{x\to \mathrm{\infty }}{lim}x\mathrm{ln}\frac{x+1}{x-1}=\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{x+1}-\frac{1}{x-1}}{\frac{-1}{{x}^{2}}}$
Find a common denominator for the two terms in the numerator, cancel fractions, and you will arrive at the desired answer of 2.
Note: Indeed, $\mathrm{ln}\frac{x+1}{x-1}\to 0$ and ${x}^{-1}\to 0$ as $x\to \mathrm{\infty }$. Therefore, we have the required form $0/0$ and L'Hopital's rule is applicable.

Do you have a similar question?