Epsilon-Delta Criterion:$ f(x) = \frac{1}{x^2 + 4}, D(f) = \Bbb R$

Question

I am new to the Epsilon-Delta Criterion and got stuck.

Epsilon-Delta Criterion: $f(x) = \frac{1}{x^2 + 4}, D(f) = \Bbb R$

\begin{align*} \left| f(x) - f(x_0) \right|&= \left|\frac{1}{x^2 + 4} - \frac{1}{x_0^2 + 4}\right| \\&= \left|\frac{x_0^2 + 4}{(x^2 + 4)(x_0^2 + 4)} - \frac{x^2 + 4}{(x_0^2 + 4)(x^2 + 4)}\right| \\&= \left|\frac{x_0^2 - x^2}{x^2 x_0^2 + 4x^2+4x_0^2+16}\right| \\&= \left|\frac{(x_0 + x) (x_0 - x)}{x^2 x_0^2 + 4x^2+4x_0^2+16}\right| \end{align*}

Because $x$ and $x_0$ are always squared in the denominator and there is no minus, the denominator as a whole is always positive. But because the numerator includes a minus, and the real numbers contain negative numbers, the numerator can be negative. $\frac{|(x_0 + x) (x_0 - x)|}{x^2 x_0^2 + 4x^2+4x_0^2+16}$

My problem is how do I determine a suitable delta?

Answer 1

First: Please juse $\LaTeX$ instead of HTML codes to style your text… you can use $\Bbb R$ to get an $\Bbb R$ or $x_0$ and $x^2$ to get $x_0$ resp. $x^2$.

Then you already have $$\left|\frac{x_0^2 - x^2}{(x_0^2+4)(x^2+4)}\right| \le \left|\frac{(x_0 + x)(x_0 -x)}{(0+4)(0+4)}\right| = \frac{1}{16}|x_0 + x|\cdot|x_0-x|$$

For $|x-x_0| < \delta$ we have $|x_0 + x| \le 2|x_0| + \delta$ hence we get $$|f(x) - f(y)| \le \frac{1}{16}|x_0 + x| \cdot |x_0 - x| \le \frac{1}{16}(2|x_0| + \delta)\delta$$

I guess you can find your $\delta$ yourself…