TITLE: show that the function $f(x)= \frac{x^2+1}{x+1}$ is continuous at the point $x_0=1$
QUESTION [5 upvotes]: Here is my attempt.
So starting with $f(x_0)=f(1)=1$ and $\epsilon >0$, I want to find $\delta >0$ such that if $|x-1|< \delta$ I get $|f(x)-f(x_0)| \leq \epsilon$.
I started studying $|f(x)-f(x_0)|$ as follows 
$|f(x)-f(x_0)|= |\frac{x^2+1}{x+1}-1|= |\frac{x^2-x}{x+1}|= |x \frac{x-1}{x+1}| \leq |\frac{x}{x+1}| \delta$
But now I cannot find a way to conclude. I want to find an upper bound for $|\frac{x}{x+1}|$ and then find an expression for $\delta$ in terms of $\epsilon$
$|f(x)-f(x_0)|= |\frac{x}{x+1}| \delta \leq \frac{(\delta +1)\delta}{2- \delta}$ 
From now on I tried to set $ \frac{(\delta +1)\delta}{2- \delta} = \epsilon$ and solve for $\delta$ but this didn't work.
I then noticed that $\frac{(\delta +1)\delta}{2- \delta} \leq \frac{(\delta +1)^2}{2- \delta}$ from where I tried solving $\frac{(\delta +1)^2}{2- \delta}=\epsilon$ but again, I failed to find an applicable solution.
I have the feeling that I am near but somehow I cannot see how to overcome the difficulty.
Thank you for your help.

REPLY [2 votes]: You could choose a $\delta$ which ensures both $|\frac{x}{x+1}| < 1$ and $\delta < \epsilon$ so you can easily extend the steps in your first line of workings.
It can be shown that $|\frac{x}{x+1}| < 1 \ \text{if and only if} \ x > -\frac 12$, and we know $ 0 < |x-1| < \delta \iff 1 - \delta < x < 1 + \delta$ and $x \neq 1$. Therefore, we want a $\delta$ such that $1 - \delta \geq -\frac 12 \iff \delta \leq \frac 32$.
We have narrowed down our requirements for $\delta$ to 

$\delta \leq \frac 32$
$\delta < \epsilon$
$\delta > 0$

I would then suggest finding a function of a real positive variable which satisfies these conditions.
An example of such $\delta$ is $\delta = \frac{\epsilon}{\epsilon+1}$.
$\textbf{Edit}$: It is fine to relax the condition $\delta < \epsilon$ to $\delta \leq \epsilon$ as shown in Cameron's answer.