What is the limit of $sqrt(4-x^2)$ as $x$ approaches $2$ ?

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Answer 1 · 2015-05-22T14:05:19

It is $0$ . (Edit: I am incorrect on this. The limit does not exist. The limit as x goes to 2 from the left, is 0.)

$lim_(xrarr2) x^2 = 4$ , so

$lim_(xrarr2) (4-x^2) = 4-4=0$ .

Therefore,

$lim_(xrarr2) sqrt(4-x^2) = sqrt(4-4)=sqrt0 =0$ .
(Edit: my error is in this step. We cannot use one sided continuity to evaluate a two sided limit.)

The graph of $sqrt(4-x^2)$ :
graph{sqrt(4-x^2) [-10, 10, -5, 5]}

I stand corrected. The limit does not exist. (I overlooked the domain issue.)

What is the limit of sqrt(4-x^2)sqrt(4-x^2) as xx approaches 22?