$[-/1\mathrm{.}04$ Points$]$

SCALCET$92\mathrm{.}2\mathrm{.}002.\frac{0}{2}$ Submissions Used

Explain what it means to say that $\underset{x{2}^{-}}{lim}f(x)=9$ and $\underset{x{2}^{+}}{lim}f(x)=6.$

As $x$ approaches $2,f(x)$ approaches $9,$ but $f(2)=6.$

As $x$ approaches $2$ from the left, $f(x)$ approaches $9.$ As $x$ approaches $2$ from the right, $f(x)$ approaches $6.$

As $x$ approaches $2,f(x)$ approaches $6,$ but $f(2)=9.$

As $x$ approaches $2$ from the right, $f(x)$ approaches $9.$ As $x$ approaches $2$ from the left, $f(x)$ approaches $6.$

In this situation is it possible that $\underset{x2}{lim}f(x)$ exists? Explain.

Yes, $f(x)$ could have a hole at $(2,9)$ and be defined such that $f(2)=6.$

Yes, $f(x)$ could have a hole at $(2,6)$ and be defined such that $f(2)=9.$

Yes, if $f(x)$ has a vertical asymptote at $x=2,$ it can be defined such that $\underset{x{2}^{-}}{lim}f(x)=9,\underset{x{2}^{+}}{lim}f(x)=6,$ and $\underset{x2}{lim}f(x)$ exists.

No$,\underset{x2}{lim}f(x)$ cannot exist if $\underset{x{2}^{-}}{lim}f(x)\underset{x{2}^{+}}{lim}f(x).$