Question

Prove the following limit. lim x → 5 3x − 8 = 7 SOLUTION 1. Preliminary analysis of the problem (guessing a value for δ). Let ε be a given positive number. We want to find a number δ such that if 0 < |x − 5| < δ then |(3x − 8) − 7| < ε. But |(3x − 8) − 7| = |3x − 15| = 3 . Therefore, we want δ such that if 0 < |x − 5| < δ then 3 < ε that is, if 0 < |x − 5| < δ then < ε 3 . This suggests that we should choose δ = ε/3. 2. Proof (showing that δ works). Given ε > 0, choose δ = ε/3. If 0 < < δ, then |(3x − 8) − 7| = = 3 < 3δ = 3 = ε. Thus if 0 < |x − 5| < δ then |(3x − 8) − 7| < ε. Therefore, by the definition of a limit lim x → 5 3x − 8 = 7.

Answer 1

$\displaystyle\lim_{x\to5}3x-8=7$

means to say that for any given $\varepsilon>0$, we can find $\delta$ such that anytime $|x-5|$ (i.e. the whenever $x$ is "close enough" to 5), we can guarantee that $|(3x-8)-7|$ (i.e. the value of $3x-8$ is "close enough" to the limit value).

What we want to end up with is

$|(3x-8)-7|=|3x-15|=3|x-5|$

Dividing both sides by 3 gives

$|x-5|$

which suggests $\delta=\dfrac\varepsilon3$ is a sufficient threshold.

The proof itself is essentially the reverse of this analysis: Let $\varepsilon>0$ be given. Then if

$|x-5|$

and so the limit is 7. QED