# Prove using the epsilon definition

I'm trying to prove the below using the $\epsilon$ definition:

$\epsilon$-definition: $\;\left|s_n-s\right| \lt \epsilon$

$\lim \limits_{n \to \infty}\frac{(-1)^n\cos \sqrt{n}}{\sqrt[3]{n}}=0$

I'm having trouble with getting the inequality into a form where I can isolate $n$

This is the best I've come up with:

$1 \lt \frac{n\epsilon}{\left(\cos\sqrt{n}\right)^3}$

• Do you mean as $n$ goes to infinity? If so, then just realize the numerator is bounded and make $n$ in the denominator large. – jdods Jul 2 '16 at 20:07

Note that $$\left| \frac { (-1)^{ n }\cos \sqrt { n } }{ \sqrt [ 3 ]{ n } } \right| =\frac { \left| (-1)^{ n } \right| \left| \cos \sqrt { n } \right| }{ \left| \sqrt [ 3 ]{ n } \right| } <\frac { 1 }{ \left| \sqrt [ 3 ]{ n } \right| } <\varepsilon$$ so $$n>\frac { 1 }{ { \varepsilon }^{ 3 } }$$ and take $${ n }_{ \varepsilon }=\left\lfloor \frac { 1 }{ { \varepsilon }^{ 3 } } \right\rfloor$$ it means when ${ n\ge n }_{ \varepsilon }$ we get $$\left| \frac { (-1)^{ n }\cos \sqrt { n } }{ \sqrt [ 3 ]{ n } } -0 \right| <\varepsilon$$