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Live4eva_2
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Could someone please give me a walkthrough of the following question(and answer)??
I really can't understand it...
lim x^2 = 9
x->3
if 0<|x-c|<delta then |f(x) - L|< epsilon
so... x^2 - 9 = (x+3)(x-3)
|x^2 - 9| = |x+3||x-3|
Here's the problem.The book states:
anyway:
if |x-3|<1 then 2<x<4 (Choosing a delta...now x lies between 2 and 4.)
|x+3|<=|x|+|3|= x+3<7 (this would be: c-delta<x<c+delta,correct?) **How do I know when to implement that ??**Why are they now using |x+3| instead of (x-3)?!?
if |x-3|<1 then |x^2 -9|< 7|x-3|
I understand that this solution basically states delta = 1/7 epsilon??And I know that the 7 is acquired through the c-delta<x<c+delta formula/method but I'm still too lost!
please clarify guys!
I really can't understand it...
lim x^2 = 9
x->3
if 0<|x-c|<delta then |f(x) - L|< epsilon
Ok,I understand that...I need x-3<delta and x^2 -9<epsilon
so... x^2 - 9 = (x+3)(x-3)
factorise...
|x^2 - 9| = |x+3||x-3|
Here's the problem.The book states:
An estimate of what??And how do I know when I must estimate??I'm guessing that it doesn't matter whether (x+3) or (x-3) is chosen due to the absolute value brackets..?We need to get an estimate for |x+3| close to 3.For convenience we'll
take within one unit of 3.
anyway:
if |x-3|<1 then 2<x<4 (Choosing a delta...now x lies between 2 and 4.)
|x+3|<=|x|+|3|= x+3<7 (this would be: c-delta<x<c+delta,correct?) **How do I know when to implement that ??**Why are they now using |x+3| instead of (x-3)?!?
if |x-3|<1 then |x^2 -9|< 7|x-3|
I understand that this solution basically states delta = 1/7 epsilon??And I know that the 7 is acquired through the c-delta<x<c+delta formula/method but I'm still too lost!
please clarify guys!
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