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QUESTION

Using l'Hopital's rule, or otherwise, evaluate
$\ds\lim_{x\to0}\left(\frac{x\sinh x}{\cosh x-1}\right)$.

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ANSWER

As $x\to0$ then $\ds\left(\frac{x\sinh x}{\cosh
x-1}\right)\rightarrow\frac{0}{0},$ an indeterminate form so using
l'Hopital's rule gives
\\ $\ds\left(\frac{x\cosh x+\sinh x}{\sinh x}\right)\rightarrow\frac{0}{0},$
also an indeterminate form.

Using l'Hopital's rule again gives

$\ds\lim_{x\to0}\left(\frac{x\sinh x+\cosh x+\cosh x}{\cosh
x}\right)=\frac{0+1+1}{1}=2$




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