Use the Sandwich/Squeeze Theorem to prove that $\displaystyle \lim_{n\rightarrow \infty} 2^{-n}\cos(n^3-n^2-13)$ exists. What is the limit?
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Originally Posted by Caity Use the Sandwich/Squeeze Theorem to prove that $\displaystyle \lim_{n\rightarrow infinity} 2^{-n}\cos(n^3-n^2-13)$ exists. What is the limit? $\displaystyle |2^{-n} \cos (n^3 -n^2 -13) | \leq 2^{-n} \to 0$
Hello, Originally Posted by Caity Use the Sandwich/Squeeze Theorem to prove that $\displaystyle \lim_{n\rightarrow infinity} 2^{-n}\cos(n^3-n^2-13)$ exists. What is the limit? Use the fact that $\displaystyle -1 \leq \cos(x) \leq 1 \quad \text{for all x}$
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...
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 $\displaystyle 2^{-n}=\frac{1}{2^n} \to 0$ $\displaystyle -2^{-n} \leq 2^{-n} \cos(\dots)\leq 2^{-n}$ Finish it off
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