$\text { Find the } x \text {-values (if any) at which the function } f ( x ) = \frac { x } { x ^ { 2 } - 100 } \text { is not continuous. }$ Which of the discontinuities are removable?
A) 10 and $- 10$, removable
B) discontinuous everywhere
C) continuous everywhere
D) 10 and $- 10$, not removable
E) 0 , removable

Question

Quizplus · Accepted Answer

The function is not continuous at $x=10$ and $x=-10$, since the denominator becomes zero at these values. These are not removable, because the limits from the left and right are not equal:
\begin{align*}
\lim_{x	o 10^-} \frac{x}{x^2-100}&=\lim_{x	o10^-}\frac{x}{(x-10)(x+10)}=-\infty \
\lim_{x	o 10^+} \frac{x}{x^2-100}&=\lim_{x	o10^+}\frac{x}{(x-10)(x+10)}=\infty \
\lim_{x	o -10^-} \frac{x}{x^2-100}&=\lim_{x	o-10^-}\frac{x}{(x-10)(x+10)}=-\infty \
\lim_{x	o -10^+} \frac{x}{x^2-100}&=\lim_{x	o-10^+}\frac{x}{(x-10)(x+10)}=\infty \
\end{align*}
Therefore, the answer is $\boxed{\mathrm{(D)}}.$.