Answer
please see step-by -step
Work Step by Step
$\displaystyle \lim_{x\rightarrow a}f(x)=L$ if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that the following is valid:
$($if $ 0 < |x-a| < \delta$ then $|f(x)-L| < \epsilon)$
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Given any $\epsilon > 0$, we want to find a $\delta > 0$ such that
$ 0 < | x-10| < \delta \ \ \Rightarrow\ \ |3-\displaystyle \frac{4}{5}x-(-5)| < \epsilon$.
Analyzing $|3-\displaystyle \frac{4}{5}x-(-5)| < \epsilon,\ \qquad(*)$
we write equivalent inequalities:
$|8-\displaystyle \frac{4}{5}x| < \epsilon$,
factor out $\displaystyle \frac{4}{5}$ on the LHS, as we aim for $| x-10|$ ...$8=\displaystyle \frac{40}{5}$
$|\displaystyle \frac{4}{5}||10-x| < \epsilon$
$|\displaystyle \frac{4}{5}||x-10| < \epsilon\qquad /\times\frac{5}{4}$
$|x-10| < \displaystyle \frac{5}{4}\epsilon$
Remember that this is equivalent to (*).
So, for any given $\epsilon > 0 $, we take $\displaystyle \delta=\frac{5}{4}\epsilon$ and
$0 < |x-10| < \delta \ \ \Rightarrow\ \ |3-\displaystyle \frac{4}{5}x-(-5)| < \epsilon$,
(the implication is one of two that comprise the equivalence)
which, by the definition of a limit, means that.
$\displaystyle \lim_{x\rightarrow 10}(3-\frac{4}{5}x)=-5$
