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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.

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

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

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