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In the function described shown below, we know that as x approaches 3, f(x) approaches 5
How close to 3 does x have to get for f(x) to differ from 5 by less than 0.1?
x is not at 3, so the distance from x to 3 can be given by lx-3l f(x) isn’t at 5, just within 0.1 of it , so l f(x)-5 l < 0.1 How much f(x) is off from the true limit (L) is indicated by the symbol € (epsilon) l f(x)-5 l < 0.1< € We want to find out how much the x can be off, δ (delta), if the y is off by €
I x – 3 I < δ I f(x)-5 I < .1 I (2x-1)-5 I < .1 I 2x-6 I < .1 2I x-3 I < .1 I x-3 I < .1/2 or .05 Since I x- 3 I is also equal to δ, it follows that δ is .05 If x is off by .05, y will be off by .1
Definition Let f be a function defined on some interval that contains a. The limit of f(x) as x approaches a is L if for every number €>0 there is a corresponding number δ>0 such that when
Example: Prove that
Homework Pg. 56 #29-30, 33-34, 37-39