Answer
$a)$ $slower$
$b)$ $slower$
$c)$ $slower$
$d)$ $faster$
$e)$ $slower$
$f)$ $slower$
$g)$ $same$
$h)$ $slower$
Work Step by Step
$a)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{x-3}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{e^{x}}$ = $0$, $slower$
$b)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{x^{3}+sin^{2}x}{e^{x}}$
= $\mathop {\lim }\limits_{n \to \infty }$$\frac{3x^{2}+2sin(x)cos(x)}{e^{x}}$
= $\mathop {\lim }\limits_{n \to \infty }$$\frac{3x^{2}+sin(2x)}{e^{x}}$
= $\mathop {\lim }\limits_{n \to \infty }$$\frac{6x+2cos(2x)}{e^{x}}$
= $\mathop {\lim }\limits_{n \to \infty }$$\frac{6-4sin(2x)}{e^{x}}$ = $0$, $slower$
$Sandwich Theorem$ $\frac{2}{e^{x}}$ $\leq$ $\frac{6-4sin(2x)}{e^{x}}$ $\leq$ $\frac{10}{e^{x}}$
$c)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{\sqrt x}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{2\sqrt{x}e^{x}}$ = $0$, $slower$
$d)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{4^{x}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$(\frac{4}{e})^{x}$ = $\infty$, $faster$
$because$ $\frac{4}{e}$ $\gt$ $1$
$e)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{(\frac{3}{2})^{x}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$(\frac{3}{2e})^{x}$ = $0$, $slower$
$because$ $\frac{3}{2e}$ $\lt$ $1$
$f)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{e^{\frac{x}{2}}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{e^{\frac{x}{2}}}$ = $0$, $slower$
$g)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{\frac{e^{x}}{2}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{{e^{x}}}{2e^{x}}$ = $\frac{1}{2}$, $same$
$h)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{\log_{10}x}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{\ln{x}}{{\ln10}e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{{\ln10}xe^{x}}$= $0$, $slower$
