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

1 Answer
Jim H
Sep 25, 2016

Always try substitution first. (it won't work for this one.) When finding a limit of a fraction and in doubt, rationalize either the numerator or denominator.

Explanation:

#sqrt(x^2-9)/(x-3)#

If we rationalize the numerator, we'll be able to factor and reduce, so that looks reasonable.

#sqrt(x^2-9)/(x-3) * sqrt(x^2-9)/(sqrt(x^2-9)) = (x^2-9)/((x-3)sqrt(x^2-9))#

# = ((x-3)(x+3))/((x-3)sqrt(x^2-9))#

# = (x+3)/sqrt(x^2-9)#

Now, as #x# approaches #3#, it must be on the left. (I am assuming that we want to stay in the real numbers.)

As #xrarr3^-#, the numerator is approaching #6# and the denominator is a positive number approaching #0#.

As #xrarr3^-#, the ratio is increasing without bound.

#lim_(xrarr3^-) sqrt(x^2-9)/(x-3) = lim_(xrarr3^-) (x+3)/sqrt(x^2-9) =oo #

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