Simple (?) limit involving sequences.

Question

"Prove that $ \lim_{n \to \infty} \, \, \left(\frac{1}{1+a_n} \right) = \frac{1}{2}$ if $\lim_{n \to \infty} a_n = 1$."

I understand the algebra, but when I get to this step:

$ |1-a_n|\left| \frac{1}{2(1+a_n)} \right| < \epsilon $

I have no idea what to do. Am I allowed to just divide the right-sided product to the epsilon side? Likewise, I don't think I can bound the right-sided product, since if $-1 < a_n < -0.5$ it explodes. I am so close to getting what I want, but don't know how to get there. Any help please?

Answer 1

Ok, here is what you do, you can choose $N$ to be large enough so that $a_n$ is say $0.5$-close to $1$ for all $n\geq N$. Then after having chosen this $N$, you can bound it as follows: $$|1-a_n||\frac{1}{2(1+a_n)}|<|1-a_n|\frac{1}{2}$$ since you are decreasing the numerator you are overall increasing the fraction. Lastly, you can choose $N'$ to be large enough (make it larger that $N$) so that $|1-a_n|<2\epsilon$, and the result follows.

Answer 2

Let $\epsilon_1=\min(\epsilon, 1/2)$.

There is an $N$ such that if $n >N$, then $|a_n-1|<\epsilon_1$.

But then if $n>N$, $$|1-a_n|\left|\frac{1}{2(1+a_n)}\right|<\epsilon.$$

Remark: You saw the issue: if we allow $a_n$ to roam to near $-1$, then we cannot control $\left|\frac{1}{1+a_n}-\frac{1}{2}\right|$. The fix is straightforward. One just picks $N$ large enough that $a_n$ is near $1$. Some version of this idea is often needed.