SOLUTION: Which of the following should the limit be evaluated using the dividing out technique? Thanks for your help. limits: https://s30.postimg.org/4vh62wiwx/math1.png

Algebra ->  Test -> SOLUTION: Which of the following should the limit be evaluated using the dividing out technique? Thanks for your help. limits: https://s30.postimg.org/4vh62wiwx/math1.png      Log On


   



Question 1081520: Which of the following should the limit be evaluated using the dividing out technique? Thanks for your help.
limits: https://s30.postimg.org/4vh62wiwx/math1.png

Found 2 solutions by jim_thompson5910, MathLover1:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Problem:
Which of the following should the limit be evaluated using the dividing out technique?

--------------------------------------------------------------------------

The answer is choice B. Here's why

If we plug x = 4 into the expression, we run into the problem having 0 in the denominator. Notice how x-4 turns into 4-4 = 0 when we replace x with 4. Luckily, we can factor the numerator to get x^2-16 = (x-4)(x+4). A pair of "(x-4)" terms will cancel leaving just x+4. At this point, we plug in x = 4 to get x+4 = 4+4 = 8 which is the limiting value.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
A.
Begin by factoring the numerator and dividing out any common factors.
lim%28x-%3E3%29_+%28%28x%5E2%2Bx-6%29%2F%28x%2B3%29%29

lim%28x-%3E3%29_+%28%28x-2%29%28x%2B3%29%2F%28x%2B3%29%29

lim%28x-%3E3%29_+%28%28x-2%29cross%28%28x%2B3%29%29%2Fcross%28%28x%2B3%29%29%29

lim%28x-%3E3%29_%28x-2%29

=%28highlight%283%29-2%29

=1

B.
lim%28x-%3E4%29_+%28%28x%5E2-16%29%2F%28x-4%29%29
lim%28x-%3E4%29_+%28%28x%5E2-4%5E2%29%2F%28x-4%29%29
lim%28x-%3E4%29_+%28%28x-4%29%28x%2B4%29%29%2F%28x-4%29%29
lim%28x-%3E4%29_+%28cross%28%28x-4%29%29%28x%2B4%29%29%2Fcross%28%28x-4%29%29%29
lim%28x-%3E4%29_+%28x%2B4%29
=+%284%2B4%29
=+8

C.
lim%28x-%3E2%29_+%28%28x%5E2-2%29%2F%28x-2%29%29...cannot be factored
Limit from the left: lim%28x-%3E2%29_+%28x%5E2+-+2%29%2F%28x+-+2%29+=%282%5E2+-+2%29%2F%282+-+2%29=+-infinity
Limit from the right: lim%28x-%3E2%29_+%28x%5E2+-+2%29%2F%28x+-+2%29+=%282%5E2+-+2%29%2F%282+-+2%29=+infinity

D.I hope you will be able to this one by yourself