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

Question

7388 views around the world

You can reuse this answer
Creative Commons License

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.)