Q:

showing that $w$ is continuous at $1$

Let $w\colon\mathbb{R}\to\mathbb{R}$. Assume $w$ is increasing.  Assume $w(1) = 2$.  Assume the sequence $2 + \frac{(−1)^n}{n}$ belongs to the image of $w$. How is $w$ continuous at $1$?  This doesn't really make sense to me.  I've seen a few proofs but can't really follow how they proved this.  Thanks!  

A:

Let's fix a small $\epsilon>0$ and find a $\delta >0$ making the following implication hold:
$$|x-1| < \delta \Rightarrow  |w(x)-2| < \epsilon .$$
Let $N \in \mathbb{N}$ be odd, and be such that $\frac{1}{N} < \epsilon$. 
Then, there is an $x_1 \in \mathbb{R}$ such that $w(x_1) = 2 - \frac{1}{N}$. (Note the $(-1)^n$ term simplifies by my specification that $N$ is odd).  By the increasingness, we must have that  $x_1 < 1$. Also, by increasingness, for all $x \in (x_1, 1)$ we know that 
$$  2 - \frac{1}{N} = w(x_1) \leq w(x) \leq w(1) = 2 .$$
There is also an $x_2 \in \mathbb{R}$ such that $w(x_2) = 2 + \frac{1}{N+1}$.  By the increasingness, for all $x \in (1,x_2)$ we know that 
$$  2 + \frac{1}{N+1} = w(x_2) \geq w(x) \geq w(1) = 2 .$$
Hence, for all $x \in (x_1,x_2)$, 
$$|w(x)-2| \leq \frac{1}{N} < \epsilon.  $$
Finally we note that since $x_1 < 1 < x_2$ we can pick a little $\delta >0$ such that $(1-\delta, 1+\delta) \subset (x_1,x_2)$, and the above inequality therefore holds for all $x \in (1-\delta, 1+\delta)$, that is for all $x \in \mathbb{R}$ satisfying $|x-1| < \delta$.

