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.
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= ???????
1 answer