Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\sinh x}}{x} \cr
& {\text{By definition }}\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sinh x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{e^x} - {e^{ - x}}}}{2}}}{x} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{2x}} \cr
& {\text{Evaluating}} \cr
& = \frac{{{e^0} - {e^{ - 0}}}}{{2\left( 0 \right)}} = \frac{{1 - 1}}{0} = \frac{0}{0} \cr
& {\text{By LHopital's Rule}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{e^x} - {e^{ - x}}} \right]}}{{\frac{d}{{dx}}\left[ {2x} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}}}}{2} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}}}}{2} = \frac{{{e^0} + {e^{ - 0}}}}{2} = \frac{{1 + 1}}{2} = 1 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sinh x}}{x} = 1 \cr} $$
