do you mean x --> -1? Otherwise, there's no problem evaluating the fraction.
x^3+1 = (x+1)(x^2-x+1)
divide out the (x+1) to get
limit = 3
or, using L'Hopital's rule,
lim (3x^2)/1 = 3
x^2(3-x) / x^2(x+4) = (3-x)/(x+4) --> 3/4
lim (16-x)/(√x-4) = lim (-1)/(1/(2√x)) = -1/(1/8) = -8
or, divide out the √x-4, since 16-x = -(√x-4)(√x+4) --> -8
Evaluate each limit. If it exists.
a) Lim x^3 + 1
x->1 -------
x + 1
b) Lim 3x^2 - x^3
x->0 ----------
x^3 + 4x^2
c) Lim 16 - x
x->16 ---------
(√x) - 4
1 answer
1 answer