Limit with Epsilon - Delta method

Question

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?

Answer 1

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!

Answer 2

$|\dfrac{5}{4x-11}-5| = |\dfrac{-20x+60}{4x-11}|=$

$20|\dfrac{x-3}{4x-11}|$.

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

$|\dfrac{5}{4x-11}-5| =20\dfrac{|x-3|}{|4x-11|} <$

$20\dfrac{|x-3|}{1/2} = 40|x-3| \lt 40\delta \le \epsilon$.

Answer 3

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.