Answer
(a) $e^{x}\geq 1+x+\frac{1}{2!}x^{2}+...+\frac{x^{n}}{n!}$ if $x\geq 0$
(b) $e>2.7$
(c) $\lim\limits_{x \to \infty} \frac{e^{x}}{x^{k}}=\infty $
Work Step by Step
(a) As we are given $f(x)=e^{x}\geq 1+x+\frac{1}{2!}x^{2}+...+\frac{x^{n}}{n!}$ if $x\geq 0$
Prove this inequality .
Consider $f(x)=e^{x}-[1+x+\frac{1}{2}x^{2}]$
Now, $f'(x)=e^{x}- (1+x)$
$f'(0)=e^{0}- 1=0$
Since, $f(x)$ is increasing on $[0,\infty)$
Thus, $e^{x}\geq 1+x+\frac{1}{2!}x^{2}+...+\frac{x^{n}}{n!}$ if $x\geq 0$
(b) Using part (a), we have
$e^{x}\geq 1+x+\frac{1}{2!}x^{2}+...+\frac{x^{n}}{n!}$ for $x\geq 0$
Take $n=4$ and $x=1$
$e^{1}\geq 1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}=2.7083$
Hence, $e>2$
(c) $f(x)=e^{x}\geq 1+x+\frac{1}{2!}x^{2}+...+\frac{x^{k}}{k!}+\frac{x^{k+1}}{(k+1)!}$ for $x\geq 0$
Then
$\frac{e^{x}}{x^{k}}\geq \frac{1}{x^{k}}+x+\frac{1}{x^{k-1}}x^{2}+...+\frac{1}{k!}+\frac{x}{(k+1)!}\geq \frac{x}{(x+1)!}$ for $x\geq 0$
Since, $\lim\limits_{x \to \infty} \frac{x}{(x+1)!}=\infty $
Hence, $\lim\limits_{x \to \infty} \frac{e^{x}}{x^{k}}=\infty $; for any $k>0$