<s>[INST] 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
 [/INST] 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 $\left| \frac{(x-1)(x+1)}{x^2+3}\right|< |x-1|$
Can you complete the task?
Your previous mistake:
$$\left| \frac{2(x^2+1)-(x^2+1)}{x^2+3}\right| < \epsilon$$
the second $x^2+1$ in the numerator should be $x^2+3$.
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