Limit with Epsilon - Delta method Prove using the $\epsilon - \delta$ definition of limits that $\lim_{x\to3} \frac{5}{4x-11} = 5$.
I know how the setup should be given $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that $|x-3| \lt \delta$ and $|\frac{5}{4x-11} - 5| \lt \epsilon$ but I can't do the computation to help me find $\delta$ can someone guide me in the right direction?

Let $\varepsilon > 0$ and define $\delta = \min\left\{\frac{\varepsilon}{4(5+\varepsilon)}, \frac14\right\}$. For $|x-3| < \delta$ we have
$$\left|5 - \frac{5}{4x-11}\right| = \left|\frac{20x-60}{4x-11}\right| \le \frac{20\left|x-3\right|}{\left|4x-11\right|} \le \frac{20\left|x-3\right|}{1 - \left|4x-12\right|} = \frac{20\left|x-3\right|}{1 - 4\left|x-3\right|} < \varepsilon$$
because
\begin{align}
\frac{20\left|x-3\right|}{1 - 4\left|x-3\right|} < \varepsilon &\iff 20\left|x-3\right| < \varepsilon(1 - 4\left|x-3\right|) \\
&\iff (20 + 4\varepsilon)\left|x-3\right| < \varepsilon\\
&\iff |x-3| < \frac{\varepsilon}{4(5+\varepsilon)}
\end{align}
which is true.
