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^2x+1)
divide out the (x+1) to get
limit = 3
or, using L'Hopital's rule,
lim (3x^2)/1 = 3
x^2(3x) / x^2(x+4) = (3x)/(x+4) > 3/4
lim (16x)/(√x4) = lim (1)/(1/(2√x)) = 1/(1/8) = 8
or, divide out the √x4, since 16x = (√x4)(√x+4) > 8
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