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
Let $\varepsilon > 0$ and choose $\delta = \sqrt{\varepsilon}$, then $|x-0|<\delta$ gives us
$$\left|\frac{1}{x^2+1}-1\right| = \left|\frac{1-(x^2+1)}{x^2+1}\right| = \frac{x^2}{x^2+1}<\frac{x^2}{1}<\delta^2=(\sqrt{\varepsilon})^2=\varepsilon.$$
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 $b$
