For the function f(x) = (x²-16)/(x-4), what is the value that f(x) approaches as x approaches 4, even though f(4) is undefined?

Factor the numerator: $\frac{x^2-16}{x-4}=\frac{(x-4)(x+4)}{x-4}=x+4$ for $x\ne4$. The function has a hole at $x=4$, and the value it approaches is $4+4=8$, choice C. Choice A ($0$) is the numerator's value at $x=4$ with no regard for the zero in the denominator. Choice B ($4$) is the $x$-value of the hole, not the height. Choice D ($16$) is the constant in the numerator. Because the factor $x-4$ cancels, there is no vertical asymptote here.