Answer
$a)$ $slower$
$b)$ $slower$
$c)$ $slower$
$d)$ $slower$
$e)$ $slower$
$f)$ $faster$
$g)$ $slower$
$h)$ $same$
Work Step by Step
$a)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{10x^{4}+30x+1}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{40x^{3}+30}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{120x^{2}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{240x}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{240}{e^{x}}$ = $0$, $slower$
$b)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{x\ln{x}-x}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{x(\frac{1}{x})+\ln{x}-1}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{\ln{x}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{xe^{x}}$ = $0$, $slower$
$c)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{\sqrt{1+x^{4}}}{e^{x}}$
= $\sqrt {\mathop {\lim }\limits_{n \to \infty }\frac{1+x^{4}}{e^{x}}}$
= $\sqrt {\mathop {\lim }\limits_{n \to \infty }\frac{4x^{3}}{e^{x}}}$
= $\sqrt {\mathop {\lim }\limits_{n \to \infty }\frac{12x^{2}}{e^{x}}}$
= $\sqrt {\mathop {\lim }\limits_{n \to \infty }\frac{24x}{e^{x}}}$
= $\sqrt {\mathop {\lim }\limits_{n \to \infty }\frac{24}{e^{x}}}$ = $\sqrt 0$ = $0$, $slower$
$d)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{(\frac{5}{2})^{x}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$(\frac{5}{2e})^{x}$ = $0$, $slower$
$because$ $\frac{5}{2e}$ $\lt$ $1$
$e)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{e^{-x}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{e^{2x}}$ = $0$, $slower$
$f)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{xe^{x}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$x$ = $\infty$, $faster$
$g)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{e^{cos(x)}}{e^{x}}$
$-1$ $\leq$ $cos(x)$ $\leq$ $1$
$\frac{e^{-1}}{e^{x}}$ $\leq$ $\frac{e^{cos(x)}}{e^{x}}$ $\leq$ $\frac{e^{1}}{e^{x}}$
$\mathop {\lim }\limits_{n \to \infty }$$\frac{e^{-1}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{e^{x+1}}$ = $0$
$\mathop {\lim }\limits_{n \to \infty }$$\frac{e^{1}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{e^{x-1}}$ = $0$
so $\mathop {\lim }\limits_{n \to \infty }$$\frac{e^{cos(x)}}{e^{x}}$ = $0$, $slower$
$h)$ $\mathop {\lim }\limits_{n \to \infty }$$\frac{e^{x-1}}{e^{x}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{e^{x-x+1}}$ = $\mathop {\lim }\limits_{n \to \infty }$$\frac{1}{e}$ = $\frac{1}{e}$, $same$
