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How could it be shown that $$\lim_{x\to0}\left[x\ln\left(\frac{e^x+1}{e^x-1}\right)\right]=\lim_{x\to\infty}\left[x\ln\left(\frac{e^x+1}{e^x-1}\right)\right]=0\quad?$$

Note that when $x=0$, we have $x=0$ (obviously) and $\ln\left(\frac{e^x+1}{e^x-1}\right)=\ln\left(\frac20\right)$ which is indeterminate and when $x\to\infty$, we have $x\to\infty$ (obviously) and $\ln\left(\frac{e^x+1}{e^x-1}\right)=\ln\left(1+\frac2{e^x-1}\right)\to\ln1=0$. So it is not possible to just multiply both.

I can't see L'Hopital working as the fraction in $\ln$ expands to a sum, not a fraction.

Here's a plot of the function in Desmos.

Any approaches?

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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    Take lim separately for x and rest of the function then in the rest of the function take lim inside log and then evaluate by l' hopital to get the answer – Pi_die_die Jul 09 '18 at 08:20

1 Answers1

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$x\ln (\frac {e^{x}+1} {e^{x}-1})=x\ln {(e^{x}+1)}-x\ln {(e^{x}-1)}$ and it is clear that $x\ln {(e^{x}+1)} \to 0$ as $x \to 0$. To find the limit of $x\ln {(e^{x}-1)}$ apply L'Hopital Rule to $\frac {\ln (e^{x}-1)} {1/x}$ and use the fact that $\frac {x^{2}} {e^{x}-1} \to 0$ as $x \to 0$ (by Another application of L'Hopital Rule). For limit as $x \to \infty $ use the fact that $\ln \frac { {(e^{x}+1)}} { {(e^{x}-1)}}=\ln (1+\frac 2 {e^{x}-1})$ and then use the fact that $\ln (1+x) $ behaves like $x$ as $x \to 0$.

Kavi Rama Murthy
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