Why can't you use substitution. Find the limit if it exists. limit (x--> -2): square root of x-2
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2 Answers

thats same as sqrt(x) as x gotu 0...anser=0

????????? substitute wot ?????????

me assume yu come from rite side so x alwaes > 0, so take root av + number
edited by

When x=2, the square root is 0, a real number, but as soon as x<2, x-2 becomes negative and the square root becomes imaginary. Where i is the imaginary square root of -1, as x approaches -2 the square root approaches 2i or i*sqrt(2-x)=i*sqrt(4) (strictly speaking it's +2i). For example, if x=1, the expression becomes +i, an imaginary number; or if x=0, we get +i*sqrt(2). Substitution is not applicable in the usual way because of this switch between real and imaginary numbers.

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