# Proving a limit exists using the definition of convergence

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

$$|\frac{\sin(n)-2\cos(n)}{3n+\cos(n)}|<\epsilon$$

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.

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 $$N=[\frac 1 {\epsilon} +\frac 1 3]+1$$