how do i find the limits of 4 -x^2/ x t 2 as x approaches -2?

Answer 1

This question is really only interesting if what you MEANT to type was:
(4 - x^2) / (x + 2)

So you want to solve:

. lim (4 - x^2) / (x + 2)
x→ -2

Of course the original expression is not defined at x = -2, because that would involve dividing by zero. However, for any real numbers x ≠ -2:
(4 - x^2) / (x + 2) = (2 - x)(2 + x) / (x + 2)

Since x ≠ -2 it is valid to divide top and bottom of the fraction by the common term (x + 2) which leaves:

(4 - x^2) / (x + 2) = (2 - x), for all x ≠ -2.

Thus,
. lim (4 - x^2)/(x + 2)
x→ -2
= lim (2 - x) = 2 - (-2) = 4 <<<<<<<<<<<
. x→ -2

Answer 2

(4-x^2)/x t 2
I'm not sure what that t is for. I'm sure it's a typo. You should add additional details and retype your question.
Assuming the t is a plus sign, you would do the following:
(4-x^2)/(x+2)
Factor 4-x^2
4-x^2 is a difference of squares.
4-x^2 = (2-x)(2+x)
2+x is the same thing as x+2, so you can now cancel the (x+2)'s.
(4-x^2)/(x+2)
= 2-x
Replace x with -2
2-(-2)
= 2+2 = 4

Answer 3

exactly as you typed it you could just plug in -2 and get 8 since it is a line, but if you mean (4-x^2)/(x+2) then the answer would be 4. You get this by factoring the top of the equation into (x+2)(-1)(x-2). You then cancel the (x+2) on top with the (x+2) on the bottom and then you plug in -2.

Answer 4

thats a toughie. too bad all the math geniuses don't use Yahoo! Answers. They're probably having a math conference out in Boring, Oregon. hahaha

Answer 5

homework question?