Answer
$2$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}} - 2\cos x}}{{x\sin x}} \cr
& {\text{Use the quotient property of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 0} \left( {{e^x} + {e^{ - x}} - 2\cos x} \right)}}{{\mathop {\lim }\limits_{x \to 0} \left( {x\sin x} \right)}} \cr
& {\text{Evaluate the limit when }}x \to 0 \cr
& = \frac{{{e^0} + {e^{ - 0}} - 2\cos \left( 0 \right)}}{{0\sin 0}} = \frac{{1 + 1 - 2}}{0} = \frac{0}{0} \cr
& {\text{The limit is an indeterminate form type }}\frac{0}{0},{\text{ so we can apply}} \cr
& {\text{the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}} - 2\cos x}}{{x\sin x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{e^x} + {e^{ - x}} - 2\cos x} \right]}}{{\underbrace {\frac{d}{{dx}}\left[ {x\sin x} \right]}_{{\text{product rule}}}}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} + 2\sin x}}{{x\cos x + \sin x}} \cr
& {\text{Evaluate the limit when }}x \to 0 \cr
& = \frac{{{e^0} - {e^{ - 0}} + 2\sin \left( 0 \right)}}{{0\cos 0 + \sin 0}} = \frac{0}{0} \cr
& {\text{Apply the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} + 2\sin x}}{{x\cos x + \sin x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{e^x} - {e^{ - x}} + 2\sin x} \right]}}{{\underbrace {\frac{d}{{dx}}\left[ {x\cos x} \right]}_{{\text{product rule}}} + \frac{d}{{dx}}\left[ {\sin x} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}} + 2\cos x}}{{\cos x - x\sin x + \cos x}} \cr
& {\text{Evaluate the limit when }}x \to 0 \cr
& = \frac{{{e^0} + {e^{ - 0}} + 2\cos \left( 0 \right)}}{{\cos \left( 0 \right) - 0\sin 0 + \cos 0}} \cr
& = \frac{{1 + 1 + 2}}{{1 - 0 + 1}} = \frac{4}{2} = 2 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}} - 2\cos x}}{{x\sin x}} = 2 \cr} $$
