Proving the Inequality of e^x Using Taylor's Theorem

[AaronEx]

Homework Statement

Show that if 0 \le x \le a, and n is a natural number, then 1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!} \le e^x \le 1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!}+\frac{e^ax^{n+1}}{(n+1)!}

Homework Equations

I used Taylor's theorem to prove e^x is equal to the LHS of the inequality plus an error term \frac{e^cx^{n+1}}{(n+1)!}, where the series is of order n centered at x=0.

The Attempt at a Solution

With the expansion of e^x in the relevant equations section, I proved that, since the error term is greater than or equal to zero, the inequality holds true. Is this correct? As for the RHS inequality, I represented e^x as the expansion again, but this time stating that the expansion is on the interval [0,a], and so I argued that the RHS was larger than e^x since e^c \le e^a. Is this true as well? Have I left anything relevant out?

 

[mfb]

Looks fine.