Calculus Limits?

Question

limit (x->4) (x^2 - 16) / [(x)^(1/2) -2] =?

Answer 1

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

To solve this limit, multiply top and bottom by the conjugate of the bottom; multiply by (x^(1/2) + 2). The denominator becomes a difference of squares as a result.

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

Now, factor x^2 - 16 as a difference of squares.

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

Cancel the common factor of (x - 4) on top and bottom.

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

Now we can safely plug in x = 4.

(4 + 4) (4^(1/2) + 2)

(8) (2 + 2)
(8)(4)

32

Answer 2

Note that x^2 - 16 is the difference of squares. Not only that, but one of its factors is also, if you use your imagination. So you can factor it twice, and you will get three factors, one of which cancels out with the entire denominator.

Answer 3

lim (x->4) [(x^2 - 16) / (√x - 2)]
= lim (x->4) [(x-4) (x+4) / (√x - 2)]
= lim (x->4) [(√x - 2)(√x + 2) (x + 4) / (√x - 2)]
= lim (x->4) (√x + 2) (x + 4)
= (√4 + 2) (4 + 4)
= 32.