the Question is
$\frac{2n+(-1)^n}{n+1}$ converges to $2$ as $n$ goes infinite
i try to arrange $|f(x)-L|<\varepsilon$ and i managed it to
$\left|\frac{(-1)^n-2}{n+1}\right|<\varepsilon$. but i can't proceed more.
how can i deal with $(-1)^n$??? the answer gives one example of delta, which is $3/\varepsilon$
Hint: make your expression simpler using inequalities, in particular the triangle inequality. $\left\lvert \frac{(-1)^n - 2}{n + 1}\right\rvert \leq \left|\frac{(-1)^n - 2}{n}\right| \leq \left|\frac{(-1)^n}{n}\right| + \left|\frac{2}{n}\right|$ Now see if you can demonstrate that the right side of the inequality is smaller than $\epsilon$ for $n > \frac{3}{\epsilon}$