How do you evaluate the limit $\frac{2 - x}{x^{2} - 4}$ $(2-x)/(x^2-4)$ as x approaches $2$ $2$ ?

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Answer 1 · 2017-01-10T20:01:03

Factor and reduce.

We'll need to know $a - b = - (b - a)$ $a-b = -(b-a)$ .

Factor and simplify:

$\frac{2 - x}{x^{2} - 4} = \frac{2 - x}{(x - 2) (x + 2)}$ $(2-x)/(x^2-4) = (2-x)/((x-2)(x+2))$

$= \frac{- (x - 2)}{(x - 2) (x + 2)}$ $= (-(x-2))/((x-2)(x+2))$

$= \frac{- 1}{x + 2}$ $= (-1)/(x+2)$ .

So the limit is $- \frac{1}{4}$ $-1/4$

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