Answer
$2$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}} - 2}}{{{e^x} - x - 1}} \cr
& {\text{Use the quotient property of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 0} \left( {{e^x} + {e^{ - x}} - 2} \right)}}{{\mathop {\lim }\limits_{x \to 0} \left( {{e^x} - x - 1} \right)}} \cr
& {\text{Evaluate the limit when }}x \to 0 \cr
& = \frac{{{e^0} + {e^{ - 0}} - 2}}{{{e^0} - 1 - 1}} = \frac{{1 + 1 - 2}}{{1 - 0 - 1}} = \frac{0}{0} \cr
& {\text{The limit is an indeterminate form type }}\frac{0}{0},{\text{ so we can apply}} \cr
& {\text{the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}} - 2}}{{{e^x} - x - 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{e^x} + {e^{ - x}} - 2} \right]}}{{\frac{d}{{dx}}\left[ {{e^x} - x - 1} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{{e^x} - 1}} \cr
& {\text{Evaluate the limit when }}x \to 0 \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{{e^x} - 1}} = \frac{{{e^0} - {e^{ - 0}}}}{{{e^0} - 1}} = \frac{0}{0} \cr
& {\text{Apply the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{{e^x} - 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{e^x} - {e^{ - x}}} \right]}}{{\frac{d}{{dx}}\left[ {{e^x} - 1} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}}}}{{{e^x}}} \cr
& {\text{Evaluate the limit when }}x \to 0 \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}}}}{{{e^x}}} = \frac{{{e^0} + {e^{ - 0}}}}{{{e^0}}} = \frac{{1 + 1}}{1} = 2 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}} - 2}}{{{e^x} - x - 1}} = 2 \cr} $$
