To formally prove that [tex]\(\lim_{{x \to -5}} \left( \frac{4}{5}x - 5 \right) = -9 \)[/tex] using [tex]\(\epsilon\)[/tex]-[tex]\(\delta\)[/tex] definition, follow these steps:
1. Statement of the Limit:
We want to show that for every [tex]\(\epsilon > 0\)[/tex], there exists a [tex]\(\delta > 0\)[/tex] such that if [tex]\(|x + 5| < \delta\)[/tex], then [tex]\(|\left(\frac{4}{5}x - 5\right) + 9| < \epsilon\)[/tex].
2. Expression Manipulation:
Begin with the expression [tex]\(|\left(\frac{4}{5}x - 5\right) + 9|\)[/tex]:
[tex]\[
|\left(\frac{4}{5}x - 5\right) + 9| = |\frac{4}{5}x - 5 + 9| = |\frac{4}{5}x + 4|
\][/tex]
Factor out [tex]\(\frac{4}{5}\)[/tex]:
[tex]\[
|\frac{4}{5}x + 4| = \left|\frac{4}{5}(x + 5)\right|
\][/tex]
Simplify the absolute value term:
[tex]\[
= \frac{4}{5} |x + 5|
\][/tex]
3. Setting the Desired Inequality:
We need:
[tex]\[
\frac{4}{5} |x + 5| < \epsilon
\][/tex]
4. Solving for [tex]\(\delta\)[/tex]:
To find [tex]\(\delta\)[/tex] as a function of [tex]\(\epsilon\)[/tex], divide both sides of the above inequality by [tex]\(\frac{4}{5}\)[/tex]:
[tex]\[
|x + 5| < \frac{5}{4} \epsilon
\][/tex]
Hence, we can choose:
[tex]\[
\delta = \frac{5}{4} \epsilon
\][/tex]
So, for any [tex]\(\epsilon > 0\)[/tex], if we choose [tex]\(\delta = \frac{5}{4} \epsilon\)[/tex], then [tex]\(|x + 5| < \delta\)[/tex] will imply that [tex]\(|\left(\frac{4}{5}x - 5\right) + 9| < \epsilon\)[/tex].
Thus, [tex]\(\delta = \frac{5}{4} \epsilon\)[/tex] formally proves the limit.
