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.

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

What we want to end up with is

[tex]|(3x-8)-7|=|3x-15|=3|x-5|<\varepsilon[/tex]

Dividing both sides by 3 gives

[tex]|x-5|<\dfrac\varepsilon3[/tex]

which suggests [tex]\delta=\dfrac\varepsilon3[/tex] is a sufficient threshold.

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

[tex]|x-5|<\delta=\dfrac\varepsilon3\implies3|x-5|=|3x-15|=|(3x-8)-7|<\varepsilon[/tex]

and so the limit is 7. QED