How do you evaluate #(x^2-14x-32) / (xsqrt(x -64)# as x approaches 16?

1 Answer
Jim H
Sep 15, 2016

Note that as #xrarr16#, the expression takes on imaginary values. (The denominator involves #sqrt(x-64)#.)

Explanation:

In a course in the calculus or real numbers, this might be a typographic error or the desired answer might be "is not defined".

Many introductory courses give definitions of limits only for real valued functions of real variables. In such a course, the domain of the expression is #(64, oo)#, so the limit at #16# is not defined.

In a course that includes imaginary complex numbers, the fact that the numerator goes to #0# while the denominator does not should be enough to conclude that the limit is #0#. (My complex analysis is weak, but I am sure that this limit is #0#, but I'd have to work to get the proof.)

Note that the numerator can be factored:

#x^2-14-32 = (x-16)(x+2)#

at which point it is easier to see that it goes to #0# as #xrarr16#.

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