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Svetach [21]
3 years ago
13

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.
Mathematics
1 answer:
Temka [501]3 years ago
8 0

\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

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