Sandwich Theorem

**Caity** · October 26th 2008, 08:27 AM

Use the Sandwich/Squeeze Theorem to prove that

$\lim_{n\rightarrow \infty} 2^{-n}\cos(n^3-n^2-13)$

exists.

What is the limit?

**ThePerfectHacker** · October 26th 2008, 08:33 AM

Originally Posted by Caity

Use the Sandwich/Squeeze Theorem to prove that

$\lim_{n\rightarrow infinity} 2^{-n}\cos(n^3-n^2-13)$

exists.

What is the limit?

$|2^{-n} \cos (n^3 -n^2 -13) | \leq 2^{-n} \to 0$

**Moo** · October 26th 2008, 08:34 AM

Hello,

Originally Posted by Caity

Use the Sandwich/Squeeze Theorem to prove that

$\lim_{n\rightarrow infinity} 2^{-n}\cos(n^3-n^2-13)$

exists.

What is the limit?

Use the fact that $-1 \leq \cos(x) \leq 1 \quad \text{for all x}$

**Caity** · October 26th 2008, 08:38 AM

I did that Moo... then I got stuck... I have a problem saying it all... I was also wondering if I needed to use a subsequence to prove it...

**Moo** · October 26th 2008, 08:40 AM

Originally Posted by Caity

I did that Moo... then I got stuck... I have a problem saying it all... I was also wondering if I needed to use a subsequence to prove it...

Note that $2^{-n}=\frac{1}{2^n} \to 0$

$-2^{-n} \leq 2^{-n} \cos(\dots)\leq 2^{-n}$

Finish it off

Thread: Sandwich Theorem