I have a proof for the limit $$ \lim_{x \to 3} \frac{x^3 - 5x + 4}{x^2 - 2} = \frac{16}{7}.$$ I am unsure about finding a constant to make the following true: $0 < |x-3| < \delta$, then $\left|f(x) - \frac{16}{7}\right| <\epsilon$. I have the needed $|x-3|$, and used the inequality to reduce to something I could work with. See the "HERE" below.
Proof:
Let $\epsilon > 0$. Then there exists a $\delta > 0$ for $0 < |x - 3| < \delta$ such that $|f(x) - \frac{16}{7}| < \epsilon$. Then $|f(x) - \frac{16}{7}| < \epsilon$ iff \begin{align*} \left|\frac{x^3 + 5x + 4}{x^2 -2} - \frac{16}{7}\right| &= \left|\frac{7x^3 - 35x + 28 - 16x^2 + 32}{7x^2 - 14}\right| = \left|\frac{7x^3 - 16x^2 - 35x + 60}{7x^2 - 14}\right| \\ &= \left|\frac{(x-3)(7x^2 + 5x - 20)}{7x^2 - 14}\right| \\ &= |x-3|\left|\frac{7x^2 + 5x - 20}{7x^2 - 14}\right|~~~\text{HERE}\\ &\leq |x-3|\left|\frac{7x^2 + 5x}{7x^2}\right| \\ &= |x-3|\left|1+\frac{5}{7x}\right|. \end{align*} We now replace the term, $|1 + \frac{5}{7x}|$, with a constant. Assume $\delta \leq 1$. Then \begin{align*} &|x-3| < \delta \leq 1 \Rightarrow -1 < |x-3| < 1 \Rightarrow 2 < |x| < 4 \Rightarrow \frac{1}{2} > |\frac{1}{x}| > \frac{1}{4} \Rightarrow \frac{5}{14} > |\frac{5}{7x}| > \frac{5}{28} \\ & \Rightarrow 1 + \frac{5}{14} > |1 + \frac{5}{7x}| > 1 + \frac{5}{28} \\ & \Rightarrow \frac{38}{28} > |1 + \frac{5}{7x}| > \frac{33}{28}. \end{align*}
It follows that $|x+3||1 + \frac{5}{7x}| < |x+3|\frac{38}{28} < \epsilon$ iff $|x+3| < \frac{28}{38}\epsilon$. Now to guarantee our assumptions made on $\delta$, choose $\delta = \min\left\{1, \frac{28}{38}\epsilon\right\}$. Therefore, if $0 < |x - 3| < \delta$, then $$\left|f(x) - \frac{16}{7}\right| < \epsilon$$