Question 179227
<font face="Times New Roman" size="+2">

What you have is an indeterminate form.  The limit as x approaches 4 is zero for both the numerator and denominator functions.  You need to use L'Hôpital's Rule which says (in part) if the limit of the numerator is zero and the limit of the denominator is zero, both as x approaches c, then the limit of the quotient as x approaches c is equal to the limit, as x approches c of the quotient of the first derivitives of the numerator and denominator functions, given that those derivitives exist.

So, take the first derivitive of *[tex \Large f(x) = 16 - x^2], *[tex \Large f'(x) = -2x], and the first derivitive of *[tex \Large g(x) = x^3 - 64], *[tex \Large g'(x) = 3x^2].  Now take the limit of the quotient as x approaches 4:

*[tex \Large \text {       } \math \lim_{x \to 4} \frac {-2x}{3x^2}]

However, the answer is NOT zero.  It is *[tex \Large -\frac {8}{48} = -\frac {1}{6} ].

John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>