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^2-3)/(3x^2-15) x<(-2)

square root of(4-x^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^2-15=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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