Answer
$5$
Work Step by Step
Consider $f(x)=\lim\limits_{x \to 1}\dfrac{x^2+3x-4}{x-1}=\lim\limits_{x \to 1} \dfrac{a(x)}{b(x)}$ and $a(1)=0, b(1)=0$
Then $a'(x)=2x+3; b'(x)=1$
Thus, $f(1)=\dfrac{0}{0}$
This shows an Inderminate form of the limit, so apply L-hospital's rule:
$\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$
Thus,
$\lim\limits_{x \to 1}\dfrac{2x+3}{1}=5$
