Answer
$a)$ $same$
$b)$ $faster$
$c)$ $same$
$d)$ $same$
$e)$ $slower$
$f)$ $faster$
$g)$ $slower$
$h)$ $same$
Work Step by Step
$a)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{x^{2}+4x}{x^{2}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{2x+4}{2x}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{2}{2}$ = $1$, $same$
$b)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{x^{5}-x^{2}}{x^{2}}$= $\mathop {\lim }\limits_{n \to \infty }$$\frac{5x^{4}-2x}{2x}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{20x^{3}-2}{2}$ = $\mathop {\lim }\limits_{n \to \infty }$${60x^{2}}$ = $\infty$, $faster$
$c)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{\sqrt {x^{4}+x^{3}}}{x^{2}}$ = $\sqrt {\mathop {\lim }\limits_{n \to \infty }\frac{{x^{4}+x^{3}}}{x^{4}}}$ = $\sqrt {\mathop {\lim }\limits_{n \to \infty }(1+\frac{1}{x})}$ =$1$, $same$
$d)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{(x+3)^{2}}{x^{2}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{2(x+3)}{2x}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{2}{2}$ = $1$, $same$
$e)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{x\ln{x}}{x^{2}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{x(\frac{1}{x})+\ln{x}}{2x}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1+\ln{x}}{2x}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{\frac{1}{x}}{2}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{2x}$ = $0$, $slower$
$f)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{2^{x}}{x^{2}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{\ln{2}(2^{x})}{2x}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{(\ln{2})^{2}(2^{x})}{2}$ = $\infty$, $faster$
$g)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{x^{3}e^{-x}}{x^{2}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{x}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{e^{x}}$ = $0$, $slower$
$h)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{8x^{2}}{x^{2}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{16x}{2x}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{16}{2}$ = $8$, $same$
