Answer
$$\eqalign{
& \left( {\bf{a}} \right)\frac{2}{3} \cr
& \left( {\bf{b}} \right)\frac{2}{3} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\bf{a}} \right)\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{{x^2} + 2x - 8}} \cr
& {\text{Evaluate the given limit without using L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{{x^2} + 2x - 8}} = \frac{{{2^2} - 4}}{{{2^2} + 2\left( 2 \right) - 8}} = \frac{0}{0} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{{x^2} + 2x - 8}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x + 2} \right)\left( {x - 2} \right)}}{{\left( {x + 4} \right)\left( {x - 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x + 2}}{{x + 4}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{x + 2}}{{x + 4}} = \frac{{2 + 2}}{{2 + 4}} = \frac{2}{3} \cr
& {\text{Evaluate using L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{{x^2} + 2x - 8}} = \mathop {\lim }\limits_{x \to 2} \frac{{\frac{d}{{dx}}\left[ {{x^2} - 4} \right]}}{{\frac{d}{{dx}}\left[ {{x^2} + 2x - 8} \right]}} = \mathop {\lim }\limits_{x \to 2} \frac{{2x}}{{2x + 2}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{2x}}{{2x + 2}} = \frac{{2\left( 2 \right)}}{{2\left( 2 \right) + 2}} = \frac{2}{3} \cr
& \cr
& \left( {\bf{b}} \right)\mathop {\lim }\limits_{x \to + \infty } \frac{{2x - 5}}{{3x + 7}} \cr
& {\text{Evaluate the given limit without using L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to + \infty } \frac{{2x - 5}}{{3x + 7}} = \frac{\infty }{\infty } \cr
& \mathop {\lim }\limits_{x \to + \infty } \frac{{2x - 5}}{{3x + 7}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{2x}}{x} - \frac{5}{x}}}{{\frac{{3x}}{x} + \frac{7}{x}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{2 - \frac{5}{x}}}{{3 + \frac{7}{x}}} \cr
& \mathop {\lim }\limits_{x \to + \infty } \frac{{2 - \frac{5}{x}}}{{3 + \frac{7}{x}}} = \frac{{2 - \frac{5}{\infty }}}{{3 + \frac{7}{\infty }}} = \frac{2}{3} \cr
& {\text{Evaluate using L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{2x - 5}}{{3x + 7}} = \mathop {\lim }\limits_{x \to 2} \frac{{\frac{d}{{dx}}\left[ {2x - 5} \right]}}{{\frac{d}{{dx}}\left[ {3x + 7} \right]}} = \mathop {\lim }\limits_{x \to 2} \frac{2}{3} = \frac{2}{3} \cr} $$
