# Proving a function is continuous using $\varepsilon -\delta$ definition

I am trying to prove the function $f(x)$ = $\frac{x^2+1}{x^2+3}$ is continous at $x = 1$.

Here is my proof so far, but I am having difficulty showing the function is bounded.

Proof:

$\forall \varepsilon$ > 0, wants $\exists \delta$ > 0, such that $\left|\frac{x^{2} + 1}{x^{3} + 3} - \frac{1}{2}\right| < \varepsilon \Leftarrow |x-1|$.

Analysis:

$$\begin{split} \left|\frac{x^{2} + 1}{x^{3} + 3} - \frac{1}{2}\right| < \varepsilon &\Leftarrow \\ \left|\frac{2(x^{2} +1) - (x^2+1)}{2(x^{3} + 3)}\right| < \varepsilon &\Leftarrow \\ \left|\frac{2x^2 + 2 - x^2 -1}{2x^2 + 6}\right| < \varepsilon &\Leftarrow \\ \left|\frac{x^2+1}{2x^2 + 6}\right|<\varepsilon &\Leftarrow \\ |x^2+1|\frac{1}{|2x^2+6|} < \varepsilon &\Leftarrow \end{split}$$

That is as far as I got on the analysis, I'm having trouble with the rest

From

$$\left| \frac{x^2+1}{x^2+3}-\frac12\right| < \epsilon$$

we have

$$\left| \frac{2(x^2+1)-x^2-3}{x^2+3}\right| < \epsilon$$

$$\left| \frac{(x-1)(x+1)}{x^2+3}\right| < \epsilon$$

Let $|x-1|< 1$, then $0<x<2$.

Hence we have $1 < x+1 < 3$ and $3< x^2+3 < 7$

Hence $\left|\frac{x+1}{x^2+3} \right|< 1$

Hence if $|x-1| < 1$ then $\left| \frac{(x-1)(x+1)}{x^2+3}\right|< |x-1|$

$$\left| \frac{2(x^2+1)-(x^2+1)}{x^2+3}\right| < \epsilon$$
the second $x^2+1$ in the numerator should be $x^2+3$.
• How did you get $x^2 +3$? – Lucius Anderson Mar 31 '18 at 4:15