Evaluating Limits!?

Question

Evaluate: lim x => 16 (X^2-14x-32) / (x√x -64)

Answer 1

Normally we can use L'Hospital's rule, but I'm going to assume that you're not allowed to use it (since these are rational functions of polynomials).

Note that x sqrt(x) is the same as x * x^(1/2), which is the same as x^(3/2).

lim [ (x^2 - 14x - 32)/(x^(3/2) - 64) ]
x -> 16

First, write x^(3/2) as [x^(1/2)]^3.

lim [ (x^2 - 14x - 32)/(x^(1/2)^3 - 64) ]
x -> 16

Now, factor the denominator as a difference of cubes.

lim [ (x^2 - 14x - 32) ] / [ (x^(1/2) - 4) (x + 4x^(1/2) + 16) ]
x -> 16

Now, factor the numerator as a quadratic.

lim [ (x - 16)(x + 2) ] / [ (x^(1/2) - 4) (x + 4x^(1/2) + 16) ]
x -> 16

Factor x - 16 as a difference of squares; it isn't immediately obvious, but split x as sqrt(x) and sqrt(x).

lim [ (x^(1/2) - 4) (x^(1/2) + 4)) (x + 2) ] / [ (x^(1/2) - 4) (x + 4x^(1/2) + 16) ]
x -> 16

Now, we have a cancellation on the top and bottom.

lim [ (x^(1/2) + 4)) (x + 2) ] / [ (x + 4x^(1/2) + 16) ]
x -> 16

And now, we can plug in x = 16 normally.

[(16^(1/2) + 4)(16 + 2)] / (16 + 4(16)^(1/2) + 16)

([4 + 4] [18]) / (16 + 4(4) + 16)

[(8)(18)] / (16 + 16 + 16)

[(8)(18)] / [48]

(18)/6

3

Your final answer should be 3.

Answer 2

as x approaches 16 y approaches 3

Answer 3

If Mr l'Hospital helps you, you'll get to know that it is 3