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## Article 13Limiting Ratios.

To ﬁnd the limit, as $u$ approaches zero, of

 $\frac{sinhu}{u},\frac{tanhu}{u},$

which are then indeterminate in form.

By eq. (14), $sinhu>u>tanhu$; and if $sinhu$ and $tanhu$ be successively divided by each term of these inequalities, it follows that

$\begin{array}{cc}1<\frac{sinhu}{u}

but when $u\doteq 0$, $coshu\doteq 1$, $sechu\doteq 1$, hence

 $\underset{u\doteq 0}{lim}\frac{sinhu}{u}=1,\underset{u\doteq 0}{lim}\frac{tanhu}{u}=1.$ (24)

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