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Prove that $$\lim_{x\to 0}\frac1{x^2+1}=1.$$

I know after adding common denominators that we get $(1-x^2)/(x^2+1)$. But I don't know of a way to bound $x$ here from bounding my $\delta$. Can I get some help please? Exam is tomorrow. Use and epsilon-delta proof

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Since the denominator converges to $1$, there is actually nothing to prove. –  Daniel Robert-Nicoud Mar 2 at 1:34
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Your function is continuous at $f(0)$, therefore $\lim _{x\rightarrow 0} f(x) = f(0)$ –  Astrum Mar 2 at 1:39
    
I see, I mis-added, thats embarrassing –  Jack Mar 2 at 1:40

2 Answers 2

Let $\varepsilon > 0$ and choose $\delta = \sqrt{\varepsilon}$, then $|x-0|<\delta$ gives us

$$\left|\frac{1}{x^2+1}-1\right| = \left|\frac{1-(x^2+1)}{x^2+1}\right| = \frac{x^2}{x^2+1}<\frac{x^2}{1}<\delta^2=(\sqrt{\varepsilon})^2=\varepsilon.$$

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I mis-added, thats embarrassing –  Jack Mar 2 at 1:41

Well let's get started by bounding $\left|{f(x) - 1}\right|$:

$$\left|{\frac{1}{x^2 + 1} - 1}\right| = \left|{\frac{x^2}{x^2 + 1}}\right| \lt x^2 $$

The last inequality was due to teh fact that the denominator is always greater than $1$.

Therefore given any $\epsilon \gt 0$ there exists $\delta (=\epsilon ^{\frac 1 2}) \gt 0 $ such that $\left|{x - 0}\right| \lt \epsilon ^{\frac 1 2} \implies \left|{\frac{1}{x^2 + 1} - 1}\right| \lt \epsilon$.

Q.E.D.

Taking the square root here is not a problem since we are dealing with strictly positive quantities. And you should also convince yourself that $a \lt b \iff \sqrt a \lt \sqrt b$ for positive values $a$ and $b$

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