Show that lim as x to 0 (sinh x)/x = 1

Show that $\displaystyle\lim_{x\to0}\frac{\sinh x}{x} = 1$.

Step 1: The form

$\sinh0 = 0$, so this is $\dfrac00$.

Step 2: The series for $\sinh x$

From the definition $\sinh x = \dfrac{e^{x}-e^{-x}}{2}$, subtract the two exponential series: \[e^{x} = 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots, \qquad e^{-x} = 1-x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\cdots\]

The even powers cancel and the odd ones double, so the factor $\tfrac12$ leaves exactly the odd terms: \[\sinh x = x+\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\cdots = \sum_{n=0}^{\infty}\frac{x^{2n+1}}{\bigl(2n+1\bigr)!}. \tag{1}\]

Step 3: Divide by $x$

For $x\neq0$, dividing (1) term by term: \[\frac{\sinh x}{x} = 1+\frac{x^{2}}{6}+\frac{x^{4}}{120}+\cdots = \sum_{n=0}^{\infty}\frac{x^{2n}}{\bigl(2n+1\bigr)!}.\]

Every term after the first carries a positive even power of $x$ and therefore vanishes in the limit.

Step 4: Verification

\[\text{By l'H\^opital: } \lim_{x\to0}\frac{\cosh x}{1} = \cosh0 = 1 \;\;\checkmark\]

\[x = 0.1:\ \frac{0.1001668}{0.1} = 1.001668, \text{ and } 1+\frac{0.01}{6} = 1.001667 \;\;\checkmark\]

\[\text{Since } \sinh'(0) = \cosh0 = 1, \text{ the limit is simply the derivative of } \sinh \text{ at } 0 \;\;\checkmark\]