Question

question 19
for $varepsilon > 0$, find $delta > 0$ necessary to prove $lim_{x \to 5} (9x - 30) = 15$.
$delta = \frac{\varepsilon}{9}$
$delta = -\frac{\varepsilon}{9}$
$delta = \frac{\varepsilon}{9}+40$
$delta = \frac{\varepsilon}{9}+5$
mark th

Explanation:

Step1: Start with epsilon condition

We need $|(9x - 30) - 15| < \varepsilon$

Step2: Simplify the absolute value

$|9x - 45| < \varepsilon$
Factor: $9|x - 5| < \varepsilon$

Step3: Isolate $|x - 5|$

$|x - 5| < \frac{\varepsilon}{9}$

Step4: Match delta definition

By definition, $|x - 5| < \delta$, so $\delta = \frac{\varepsilon}{9}$

Answer:

$\boldsymbol{\delta = \frac{\varepsilon}{9}}$ (corresponding to the first option)