10. [-/0.56 Points] DETAILS SCALCET9 2.4.AE.002.
EXAMPLE 2
Video Example
Prove the following limit.
$\lim_{x \to 7} (3x - 6) = 15$
SOLUTION
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 if $0 < |x - 7| < \delta$ then $|(3x - 6) - 15| < \epsilon$.
But $|(3x - 6) - 15| = |3x - 21| = 3|x - 7|$. Therefore, we want $\delta$ such that
if $0 < |x - 7| < \delta$ then $3|x - 7| < \epsilon$
that is, if $0 < |x - 7| < \delta$ then $|x - 7| < \frac{\epsilon}{3}$
This suggests that we should choose $\delta = \frac{\epsilon}{3}$.
2. Proof (showing that $\delta$ works). Given $\epsilon > 0$, choose $\delta = \frac{\epsilon}{3}$. If $0 < |x - 7| < \delta$, then we get the following.
$|(3x - 6) - 15| = |3x - 21|$
$= 3|x - 7|$
$< 3$\delta$
$= 3(\frac{\epsilon}{3}) = \epsilon$
Thus, if $0 < |x - 7| < \delta$ then $|(3x - 6) - 15| < \epsilon$.
Therefore, by the definition of a limit, we get the following.
$\lim_{x \to 7} (3x - 6) = 15$
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