This problem says if n>=3, derive the inequality -x^2 <= x^n <= x^2 for -1 < x < 1. Not really sure how to even start this...
We proved that the inequality is true for n=3. Now, suppose that is true for n
$\displaystyle -x^2\leq x^n\leq x^2$ (hypothesis)
Now
$\displaystyle -1\leq x \leq 1$ multiply by $\displaystyle x^n$ both sides
$\displaystyle -x^n \leq x^{n+1} \leq x^n$
Using the hypothesis
$\displaystyle -x^2\leq -x^n \leq x^{n+1} \leq x^n \leq x^2 \Rightarrow -x^2\leq x^{n+1}\leq x^2 \quad \Box$
Hay Cat here is what is going on.
$\displaystyle -1<x<1\text{ if and only if }|x|<1$.
Now recall this $\displaystyle \left| x \right|\left| {x^n } \right| = \left| {x^{n + 1} } \right|$.
We can say the $\displaystyle x\ne 0$. For if it were there is nothing to prove.
$\displaystyle \left| x \right| < 1 \Rightarrow \left| {x^2 } \right| < \left| x \right| < 1 \Rightarrow \left| {x^3 } \right| < \left| {x^2 } \right| < \left| x \right| < 1$