Prove $\frac{2n+\sin(n)}{n+2}$ converges to 2.

Question

Here's my attempt.

Let $\epsilon >0.$ Then $N \geq \frac{5}{\epsilon}$, with $n \geq N \implies$

$| \frac{2n+\sin(n)}{n+2} -2|=| \frac{2n+\sin(n)-2n-4}{n+2}|=| \frac{\sin(n)-4}{n+2}|= \frac{|\sin(n)-4|}{n+2} \leq \frac{5}{n+2} \leq \frac{5}{n} \leq \epsilon$

Answer 1

Good job!

If you don't have to prove from definition:

\begin{align} \lim_{n \to \infty} \frac{2n + \sin n}{n+2} = \lim_{n \to \infty } \frac{2+ \frac{\sin(n)}{n}}{1+\frac{2}{n}}=2 \end{align}

Answer 2

Alternatively, $$ \frac{2n-1}{n+2} \le \frac{2n+\sin(n)}{n+2} \le \frac{2n+1}{n+2} $$ and both bounds go to $2$ as $n \to \infty$.

Answer 3

Alternatively,

$$\frac{2n+\sin n}{n+2}=2+\frac{\sin n-4}{n+2}$$

and the numerator of the fraction is bounded while the denominator increases to infinity.