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> Prove that the function $\displaystyle f(x)=\frac{x}{x^2+1}$ is continuous at $x=-1$. You should give a proof that is directly based on the definition of continuity.
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I saw many similar questions on this website, but none of them gave me a satisfactory answer. I more or less understand how epsilon-delta proofs work but my problem lies more in the algebraic manipulations probably. So here's my "attempt".

Definition of continuity; for $f:A \to \mathbb{R}$

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> $$\forall \varepsilon >0 \ \exists\delta>0\ \forall x \forall x\_0 \in A:|x-x\_0|<\delta \Rightarrow |f(x)-f(x\_0)|<\varepsilon$$
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So from what I understand, the procedure for epsilon-delta proofs are as follows:  

1. Plug the point of interest $x\_0$ into $|f(x)-f(x\_0)|<\varepsilon$  

2. Write $x$ in terms of $\varepsilon$  

3. Then plug $x$ into $|x-x\_0|<\delta$  

4. Find $\delta$ in terms of $\varepsilon$. Done 

At step 1, we have $\left|\dfrac{x}{x^2+1}+\dfrac{1}{2}\right|<\varepsilon$. Step 2 is where I'm stuck because of the 2nd degree polynomial in the denominator. First, I did this: $\left|\dfrac{x}{x^2+1}\right|<\left|\dfrac{x}{x^2+1}+\dfrac{1}{2}\right|<\varepsilon$. Then, I wasn't too sure about what to do, I thought about getting rid of the "$1$", but that would screw up my inequality. I thought about factoring $x^2+1$, but it doesn't have any real roots. I tried inverting the fraction, but that led me to nowhere. And I've already exhausted the tools I have. **How do I proceed?**

Apparently, epsilon-delta proofs are the most basic building blocks of analysis. And I'm already failing, this is both embarrassing and depressing...