A sequence ($a_n$) converges to a real number $a$ if, for every positive number $\epsilon$, there exists an $N\in \mathbb{N}$ such that whenever $n \geqslant N$ it follows that $|a_n-a| < \epsilon$

Use this to prove that lim$\frac{6n+\sin(n)}{3n+\cos(n)}=2$

Can somebody help me with this because I do not understand how to deal with the cosine and sine in the function

so we have $\frac{6n+\sin(n)}{3n+\cos(n)}-2<\epsilon$

this can be written to


I could use the triangle inequality

but i do not know how to continou i don't even know if this is a right step.


2 Answers 2


Use the fact that$$\bigl\lvert\sin(n)-2\cos(n)\bigr\rvert\leqslant3$$together with the fact that$$\bigl\lvert3n+\cos(n)\bigr\rvert\geqslant3n-1.$$


$|\frac {\sin n-2\cos n|} {3n+\cos n|} \leq \frac 3 {3n-1} <\epsilon$ if $n >\frac 1 {\epsilon} +\frac 1 3$. Take $N=[\frac 1 {\epsilon} +\frac 1 3]+1$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.