SOLUTION 1: $$\displaystyle{ \lim_{x \rightarrow 1} \frac{ x^2-1 }{ x^2+3x-4 } } = \displaystyle{ \frac{ (1)^2-1 }{ (1)^2+3(1)-4 } } = \frac{"0"}{0}$$
(Apply Theorem 1 for l'Hopital's Rule. Differentiate top and bottom separately.)
$$= \displaystyle{ \lim_{x \rightarrow 1} \frac{ 2x-0 }{ 2x+3-0 } }$$ $$= \displaystyle{ \lim_{x \rightarrow 1} \frac{ 2x }{ 2x+3 } }$$ $$= \displaystyle{ \frac{ 2(1)-0 }{ 2(1)+3 } }$$ $$= \displaystyle{ \frac{ 2 }{ 5 } }$$