Answer
$$ - 5$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{5 - 5\cos x}}{{{e^x} - x - 1}} \cr
& {\text{evaluating the limit, we get:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{5 - 5\cos x}}{{{e^x} - x - 1}} = \frac{{5 - 5\cos \left( 0 \right)}}{{{e^0} - 0 - 1}} = \frac{{5 - 5}}{{1 - 1}} = \frac{0}{0} \cr
& {\text{the limit is }}\frac{0}{0}{\text{}}{\text{, so we can apply l'Hopital's rule}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{d/dx\left( {5 - 5\cos x} \right)}}{{d/dx\left( {{e^x} - x - 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{ - 5\sin x}}{{{e^x} - 1}} \cr
& {\text{evaluating the limit, we get:}} \cr
& = \frac{{ - 5\sin \left( 0 \right)}}{{{e^0} - 1}} \cr
& = \frac{0}{{1 - 1}} = \frac{0}{0} \cr
& {\text{the limit is still }}\frac{0}{0}{\text{, so}}{\text{ we apply l'Hopital's rule again}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{d/dx\left( { - 5\sin x} \right)}}{{d/dx\left( {{e^x} - 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{ - 5\cos x}}{{{e^x}}} \cr
& = \frac{{ - 5\cos \left( 0 \right)}}{{{e^0}}} \cr
& = - 5 \cr
& {\text{then}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{5 - 5\cos x}}{{{e^x} - x - 1}} = - 5 \cr} $$
