# Thread: Limit confusion

1. ## Limit confusion

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...

2. $-1 \leq x \leq 1$ multiplying both sides by x^2

$-x^2 \leq x^3 \leq x^2$

I think that you can prove the general result for $n\geq 3$ using induction

3. Originally Posted by felper
$-1 \leq x \leq 1$ multiplying both sides by x^2

$-x^2 \leq x^3 \leq x^2$

I think that you can prove the general result for $n\geq 3$ using induction
Ok I'm kind of stuck, it's been a while since I've used induction. I'm at the point where $-x^2 \leq x^{(n+1)} \leq x^2$. I'm thinking about dividing by through by x...

4. We proved that the inequality is true for n=3. Now, suppose that is true for n

$-x^2\leq x^n\leq x^2$ (hypothesis)

Now

$-1\leq x \leq 1$ multiply by $x^n$ both sides

$-x^n \leq x^{n+1} \leq x^n$

Using the hypothesis

$-x^2\leq -x^n \leq x^{n+1} \leq x^n \leq x^2 \Rightarrow -x^2\leq x^{n+1}\leq x^2 \quad \Box$

5. Originally Posted by felper
$-x^2\leq -x^n \leq x^{n+1} \leq x^n \leq x^2 \Rightarrow -x^2\leq x^{n+1}\leq x^2 \quad \Box$
How do you know $-x^2\leq -x^n$

6. Hay Cat here is what is going on.
$-1.
Now recall this $\left| x \right|\left| {x^n } \right| = \left| {x^{n + 1} } \right|$.
We can say the $x\ne 0$. For if it were there is nothing to prove.
$\left| x \right| < 1 \Rightarrow \left| {x^2 } \right| < \left| x \right| < 1 \Rightarrow \left| {x^3 } \right| < \left| {x^2 } \right| < \left| x \right| < 1$

7. Ah yes, thanks for the help.