F(-2) = (4+4+4)/(-8+8)
= 4/0
So, it will be infinity, but plus or minus?
since F(x) = ((x-1)^2 + 3)/(x^3+8), if x is slightly more than -2, the numerator is positive and the denominator is positive, so the limit is plus infinity.
Determine the one sided limit of f(x) s x approaches -2 from the right side.
F(X) = (x^2-2x+4)/(x^3+8).
I do not know where to begin. Should I plug in the one side limit?
2 answers
x²+4x+4=(x+2)(x+2)
x³+8=(x+2)(x²-2x+4)
(x²+4x+4)/(x³+8)=
(x+2)(x+2)/(x+2)(x²-2x+4)
(x+2)/(x²-2x+4)
Lim((x²+4x+4)/(x³+8)) =
Lim((x+2)/(x²-2x+4))=
(-2+2)/((-2)²-2(-2)+4)=0/4=0
Hence Lim((x²+4x+4)/(x³+8)) as x approaches - 2 from the right =0
x³+8=(x+2)(x²-2x+4)
(x²+4x+4)/(x³+8)=
(x+2)(x+2)/(x+2)(x²-2x+4)
(x+2)/(x²-2x+4)
Lim((x²+4x+4)/(x³+8)) =
Lim((x+2)/(x²-2x+4))=
(-2+2)/((-2)²-2(-2)+4)=0/4=0
Hence Lim((x²+4x+4)/(x³+8)) as x approaches - 2 from the right =0
1 answer