# Prove that the function $f(x)=\frac{x}{x^2+1}$ is continuous at $x=-1$.

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.

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 $$f:A \to \mathbb{R}$$

$$\forall \varepsilon >0 \ \exists\delta>0\ \forall x \forall x_0 \in A:|x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\varepsilon$$

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...

Hints: If $$|x+1| <\delta$$ where $$0 <\delta <1$$ then $$x<-1+\delta$$ and this implies $$x^{2} >(-1+\delta)^{2}$$ .Hence $$|\frac x {x^{2}+1}+\frac 1 2|=\frac {(x+1)^{2}} {2(x^{2}+1)}<\frac {\delta^{2}} {2(1+ (-1+\delta)^{2})}$$. Can you proceed?

A further hint: If you choose $$\delta$$ to be $$<\frac 1 2$$ then $$(-1+\delta)< -\frac 1 2$$ which implies $$(-1+\delta)^{2} >\frac 1 4$$.

• @Sckizel I have added one more hint. Feb 2 '20 at 5:39
• @Sckizel Take $\delta <\sqrt {\epsilon} \sqrt {1+\frac 1 4}$. Feb 2 '20 at 6:26
• Your desired inequality is $$(*) \quad \left|\dfrac{x}{x^2+1}+\dfrac{1}{2}\right|<\varepsilon$$ The inequality that has been proved, assuming that $|x+1|<\delta$ for some unknown value $\delta$, is $$(**) \quad \left|\frac x {x^{2}+1}+\frac 1 2\right|<\frac {\delta^{2}} {2(1+ (-1+\delta)^{2})}$$ Now ask yourself the question: What further inequality, in addition to $(**)$ implies $(*)$? Feb 2 '20 at 15:18
• To put this more abstractly: If you desire to prove $$(*) \quad a<c$$ and you know that this is true $$(**) \quad a<b$$ what further inequality, in addition to $(**)$, implies $(*)$? Feb 2 '20 at 15:19
• Your goal is to find a value of $\delta$. By setting up that inequality and solving for $\delta$, you've found it. Feb 2 '20 at 21:47

$$f(x)$$ is continuous at $$f(-1)=-1/2$$ is finite and euals both the right and the left limits: $$\lim_{x\rightarrow -1^+} f(x)= \lim_{h \rightarrow 0} \frac{-1+h}{1+(-1+h)^2}=-\frac{1}{2}$$ $$\lim_{x\rightarrow -1^-} f(x)= \lim_{h \rightarrow 0} \frac{-1-h}{1+(-1-h)^2}=-\frac{1}{2}$$

• Thank you for your answer, but this is not what I need. The problem asks explicitly to use the epsilon-delta definition of continuity. Feb 2 '20 at 5:21