How do you find the limit of #x/sqrt(x^2-9)# as #x->3^-#?

1 Answer
Jim H
Dec 7, 2016

The limit does not exist.

Explanation:

If we are staying in the real numbers, then the domain of the function is #(-oo,-3) uu (3,oo)# so we cannot approach positive 3 from the left.

If we are allowed to use imaginary numbers, then the limit still does not exist, because as x approaches 3 from the left, the numerator goes to 3 and the denominator goes to #0# (through purely imaginary values).

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