Posted by **Jessica** on Thursday, June 21, 2012 at 1:43pm.

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

- Calculus -
**Steve**, Thursday, June 21, 2012 at 4:08pm
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

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