Answer
$${\text{Limit is not in the form }}\frac{0}{0}{\text{ or }}\frac{\infty }{\infty }$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} + 4x + 1}}{{{x^2} - x - 2}} \cr
& = \mathop {\lim }\limits_{x \to 2} \frac{{6x + 4}}{{2x - 1}}{\text{ }}\left( {{\text{Step 1}}} \right) \cr
& = \mathop {\lim }\limits_{x \to 2} \frac{6}{2}{\text{ }}\left( {{\text{Step 2}}} \right) \cr
& {\text{From the first step L'hopital's rule is used incorrectly}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} + 4x + 1}}{{{x^2} - x - 2}} = \frac{{3{{\left( 2 \right)}^2} + 4\left( 2 \right) + 1}}{{{{\left( 2 \right)}^2} - \left( 2 \right) - 2}} = \frac{{21}}{0} \cr
& {\text{The result is not in the form }}\frac{0}{0},\frac{\infty }{\infty },{\text{ so the rule}} \cr
& {\text{is not applicable}}{\text{.}} \cr} $$
