Posted by Tim on Friday, February 28, 2014 at 4:52pm.
I did some of them so can you help me with the rest.
Consider the piecewise functon f(x) =
(4x^23)/(3x^215) x<(2)
square root of(4x^2) (2)<x<(2)
5 x>2
lim x>2^ f(x)= ??????
lim x>2^+ f(x)= 0
f(2)= 0
f(4) = 5
list the values of x for which f(x) is not continues= ???????

Calc Help Please  Steve, Friday, February 28, 2014 at 5:04pm
from the left, the first definition shows the limit is 13/3
from the right, the second definition applies, and the limit = 0
Naturally, f is not continuous where 3x^215=0, or x = ±√5
Also, at x = 2, since the limit does not equal the definition.
Also at x=2, since f(2)=0, but the limit on the right is 5.
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