Question $14(1$ point$)$

If the $lm,f(x)$ exists and the $lm,$dots,$f(x)$ exists, then the following is true:

$\underset{x1}{lim}f(x)$ also exists

The function is continumes at $x=0$

$\underset{x{0}^{-}}{lim}f(x)=\underset{x{0}^{-}}{lim}f(x)$

The function is discontinuous if $f(x)$ is undefined at $x=0$ and the lim ${?}_{x{0}^{+}}f(x)=\underset{x{0}^{-}}{lim}f(x)$