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

Sep 15, 2016

Note that as $x \rightarrow 16$, 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 $\left(64 , \infty\right)$, 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:

${x}^{2} - 14 - 32 = \left(x - 16\right) \left(x + 2\right)$

at which point it is easier to see that it goes to $0$ as $x \rightarrow 16$.