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?
 A: I'll provide a short sketch
$$ \left|\frac{5}{4x-11} - 5\right| = \left|\frac{5}{4x-11} - \frac{20x - 55}{4x - 11}\right| = \left|\frac{-20x + 60}{4x - 11}\right| < \epsilon $$
which is equivalent with
$$ |x-3| < \epsilon |4x - 11|/20 $$
Now try to bound $|4x-11|$, compute $\delta$ and revert the argument! Moreover, don't forget that the function is undefined in $11/4$.
I hope this helps!
A: $|\dfrac{5}{4x-11}-5| = |\dfrac{-20x+60}{4x-11}|=$
$20|\dfrac{x-3}{4x-11}|$.
Consider $|x-3| <1/8$, then
$-1/8< x-3 <1/8$,  or $-1/2< 4x -12 <1/2$, $1/2< 4x -11 <3/2.$
Let $\epsilon >0$ be given.
Choose $\delta = \min(1/8,\epsilon/(40))$.
Then $|x-3|< \delta$ implies
$|\dfrac{5}{4x-11}-5| =20\dfrac{|x-3|}{|4x-11|} <$
$20\dfrac{|x-3|}{1/2} = 40|x-3| \lt 40\delta \le \epsilon$.
A: 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.
