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 , we can find such that anytime (i.e. the whenever is "close enough" to 5), we can guarantee that (i.e. the value of is "close enough" to the limit value).
What we want to end up with is
Dividing both sides by 3 gives
which suggests is a sufficient threshold.
The proof itself is essentially the reverse of this analysis: Let be given. Then if