Prove that limx→14x = 4 by finding a general
formula for finding δ given ε
|4x - 4| < ε
4|x - 1| < ε
|x - 1| < ε/4 = δ
Now should you show | x - 1 | < ε/4 implies |4x - 4| < ε?
Technically that is what the definition of “limit” calls for
But in this case (and many others) the algebra you did to
rewrite the inequality involving ε into one defining
δ is reversible, and so you needn’t explicitly
show that | x - 1 | < ε/4 implies |4x - 4| < ε
Prove that limx→0x3 = 0
| x3 - 0 | < ε
| x3| < ε
-ε < x3 < ε
(-ε)1/3 < x < ε1/3
-(ε1/3) < x < ε1/3 (because cube root is odd)
(“Odd function” f(x) is one where f(-x) = -f(x))
|x| < ε1/3
|x - 0| < ε1/3
Next
Limits miscellany
“Treasure hunt” (for chocolate) in sections 2.4 - 2.6
What, precisely, does it take to make a function continuous at
x = c? On an interval?