3. [-/1 Points]
DETAILS
SCALCET8 2.2.002.
Explain what it means to say that
$\lim_{x \to 3^{-}} f(x) = 7$ and $\lim_{x \to 3^{+}} f(x) = 8$.
As $x$ approaches 3 from the right, $f(x)$ approaches 7. As $x$ approaches 3 from the left, $f(x)$ approaches 8.
As $x$ approaches 3, $f(x)$ approaches 7, but $f(3) = 8$.
As $x$ approaches 3 from the left, $f(x)$ approaches 7. As $x$ approaches 3 from the right, $f(x)$ approaches 8.
As $x$ approaches 3, $f(x)$ approaches 8, but $f(3) = 7$.
In this situation is it possible that $\lim_{x \to 3} f(x)$ exists? Explain.
Yes, $f(x)$ could have a hole at $(3, 7)$ and be defined such that $f(3) = 8$.
Yes, $f(x)$ could have a hole at $(3, 8)$ and be defined such that $f(3) = 7$.
Yes, if $f(x)$ has a vertical asymptote at $x = 3$, it can be defined such that $\lim_{x \to 3^{-}} f(x) = 7$, $\lim_{x \to 3^{+}} f(x) = 8$, and $\lim_{x \to 3} f(x)$ exists.
No, $\lim_{x \to 3} f(x)$ cannot exist if $\lim_{x \to 3^{-}} f(x) \neq \lim_{x \to 3^{+}} f(x)$.
