Show that $1 + \frac{x}{1!} + \dots + \frac{x^n}{n!} \leq e^x \leq 1 + \frac{x}{1!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{e^{a}x^n}{n!}$ Show that if $0 \leq x \leq a$ and $n \in \mathbb{N}$, then $$1 + \frac{x}{1!} + \dots + \frac{x^n}{n!} \leq e^x \leq 1 + \frac{x}{1!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{e^{a}x^n}{n!}$$
Let $$e^x = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{x^n}{n!} + \frac{x^{n+1}}{(n+1)!}e^x$$
where $e^x$ is expanded by Taylor's theorem, but I am not sure if I am correctly applying Taylor's theorem because it isn't clear to me where the right side of the inequality comes from. 
How does Taylor's theorem get used to complete the inequality? Is that even the right approach to this problem?


By Taylor's theorem,
$$
e^x  = 1 + x + \frac{{x^2 }}{{2!}} +  \cdots  + \frac{{x^{n - 1} }}{{(n - 1)!}} + \frac{{x^n }}{{n!}}e^c 
$$
for all $x\geq 0$ with a suitable $0\leq c\leq x$. Now just note that if $0\leq x\leq a$ then
$$
1 \leq e^c \leq e^x \leq e^a .
$$
