Answer
$$\left( {\text{a}} \right)0,{\text{ }}\left( {\text{b}} \right) - \frac{2}{3},{\text{ }}\left( {\text{c}} \right) - \infty $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\mathop {\lim }\limits_{x \to \infty } \frac{{5 - 2{x^{3/2}}}}{{3{x^2} - 4}} \cr
& {\text{Divide both the numerator and the denominator by }}{x^2} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{5 - 2{x^{3/2}}}}{{3{x^2} - 4}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{{{x^2}}} - \frac{{2{x^{3/2}}}}{{{x^2}}}}}{{\frac{{3{x^2}}}{{{x^2}}} - \frac{4}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{{{x^2}}} - \frac{2}{{{x^{1/2}}}}}}{{3 - \frac{4}{{{x^2}}}}} \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{{{x^2}}} - \frac{2}{{{x^{1/2}}}}}}{{3 - \frac{4}{{{x^2}}}}} = \frac{{\frac{5}{\infty } - \frac{2}{\infty }}}{{3 - \frac{4}{\infty }}} = \frac{{0 - 0}}{{3 - 0}} = 0 \cr
& \cr
& \left( {\text{b}} \right)\mathop {\lim }\limits_{x \to \infty } \frac{{5 - 2{x^{3/2}}}}{{3{x^{3/2}} - 4}} \cr
& {\text{Divide both the numerator and the denominator by }}{x^{3/2}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{5 - 2{x^{3/2}}}}{{3{x^{3/2}} - 4}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{{{x^{3/2}}}} - \frac{{2{x^{3/2}}}}{{{x^{3/2}}}}}}{{\frac{{3{x^{3/2}}}}{{{x^{3/2}}}} - \frac{4}{{{x^{3/2}}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{{{x^{3/2}}}} - 2}}{{3 - \frac{4}{{{x^{3/2}}}}}} \cr
& {\text{Evaluate the limit}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{{{x^{3/2}}}} - 2}}{{3 - \frac{4}{{{x^{3/2}}}}}} = \frac{{0 - 2}}{{3 - 0}} = - \frac{2}{3} \cr
& \cr
& \left( {\text{c}} \right)\mathop {\lim }\limits_{x \to \infty } \frac{{5 - 2{x^{3/2}}}}{{3x - 4}} \cr
& {\text{Divide both the numerator and the denominator by }}{x^{3/2}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{5 - 2{x^{3/2}}}}{{3x - 4}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{{{x^{3/2}}}} - \frac{{2{x^{3/2}}}}{{{x^{3/2}}}}}}{{\frac{{3x}}{{{x^{3/2}}}} - \frac{4}{{{x^{3/2}}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{{{x^{3/2}}}} - 2}}{{\frac{3}{{{x^{1/2}}}} - \frac{4}{{{x^{3/2}}}}}} \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{{{x^{3/2}}}} - 2}}{{\frac{3}{{{x^{1/2}}}} - \frac{4}{{{x^{3/2}}}}}} = \frac{{\frac{5}{\infty } - 2}}{{\frac{3}{\infty } - \frac{4}{\infty }}} = \frac{{ - 2}}{0} = - \infty \cr} $$
