Answer
$$5$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + 3x - 4}}{{x - 1}} \cr
& {\text{Evaluating, we get:}} \cr
& = \frac{{{{\left( 1 \right)}^2} + 3\left( 1 \right) - 4}}{{1 - 1}} = \frac{{4 - 4}}{{1 - 1}} = \frac{0}{0} \cr
& {\text{Using l'Hopital's rule}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left( {{x^2} + 3x - 4} \right)}}{{\frac{d}{{dx}}\left( {x - 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{2x + 3}}{1} \cr
& {\text{Evaluating the limit, we get:}} \cr
& = 2\left( 1 \right) + 3 \cr
& = 5 \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + 3x - 4}}{{x - 1}} = 5 \cr} $$
