Prove the following limit.
$\lim_{x \to 3} (5x - 8) = 7$
1. Preliminary analysis of the problem (guessing a value for $\delta$).
Let $\epsilon$ be a given positive number. We want to find a number $\delta$ such that
$| (5x - 8) - 7 | < \epsilon$ whenever $0 < | x - 3 | < \delta$.
But
$| (5x - 8) - 7 | = | 5x - 15 | = 5 | x - 3 |
Therefore, we want $\delta$ such that
$5 | x - 3 | < \epsilon$ whenever $0 < | x - 3 | < \delta$,
that is,
$| x - 3 | < \frac{\epsilon}{5}$ whenever $0 < | x - 3 | < \delta$
This suggests that we should choose $\delta = \epsilon/5$.
2. Proof (showing that $\delta$ works).
Given $\epsilon > 0$, choose $\delta = \epsilon/5$. If $0 < | x - 3 | < \delta$, then
$| (5x - 8) - 7 | = | 5x - 8 |
= 5 | x - 3 |
< 5 \delta
= 5 (\frac{\epsilon}{5})
= \epsilon$.
Thus if $0 < | x - 3 | < \delta$, then $| (5x - 8) - 7 | < \epsilon$.
