## Calculus 8th Edition

(a) $e^{x}\geq 1+x+\frac{1}{2!}x^{2}+...+\frac{x^{n}}{n!}$ if $x\geq 0$ (b) $e>2.7$ (c) $\lim\limits_{x \to \infty} \frac{e^{x}}{x^{k}}=\infty$
(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 $k>0$