Answer
$${\text{Limit is not in the form }}\frac{0}{0}{\text{ or }}\frac{\infty }{\infty }$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } x\cos \left( {\frac{1}{x}} \right) \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\cos \left( {1/x} \right)}}{{\left( {1/x} \right)}}{\text{ }}\left( {{\text{Step 1}}} \right) \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\left[ { - \sin \left( {1/x} \right)} \right]\left( {1/{x^2}} \right)}}{{ - \left( {1/{x^2}} \right)}}{\text{ }}\left( {{\text{Step 2}}} \right) \cr
& = 0{\text{ }}\left( {{\text{Step 3}}} \right) \cr
& {\text{From the step 2 the L'Hopital's rule is used incorrectly}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{\cos \left( {1/x} \right)}}{{\left( {1/x} \right)}} = \frac{{\cos \left( {1/\infty } \right)}}{{1/\infty }} = \frac{1}{0} = \infty \cr
& {\text{The result is not in the form }}\frac{0}{0},\frac{\infty }{\infty },{\text{ }} \cr
& {\text{then the L'Hopital's Rule does not apply}}{\text{.}} \cr} $$
