# Question #bf3b9

Sep 2, 2016

Yes and no,

#### Explanation:

If f(a) is undefined, then it could be that f(a) doesn't make any sense. Example log(0). Log(x) can be continuously traveled to reach infinity at 0. Can you be more specific as to the kind of example you had in mind?

Sep 2, 2016

It might, if you specify it correctly.

If you had a vertical asymptote at $x = a$ that is like the one in $f \left(x\right) = \frac{1}{x - a}$, then the limit from each side is:

${\lim}_{x \to {a}^{+}} \frac{1}{x - a} = \infty$

${\lim}_{x \to {a}^{-}} \frac{1}{x - a} = - \infty$

However, if you choose the limit such that you approach $x = a$ from both sides at the same time:

${\lim}_{x \to a} \frac{1}{x - a} = \frac{1}{0} = \text{undefined}$

And that would be because from the left and from the right, the limits are each opposite in sign.

If you specify approaching from both sides, it is implied that the limit is "both" $\infty$ and $- \infty$, which is impossible, because a given limit can't correspond to two values at the same time.

Alternatively, if you had a removable discontinuity, you might have a function $f \left(a\right)$ that is undefined, which has a limit from a non-specific side:

$f \left(x\right) = \left\{\begin{matrix}x & x < 1 \\ {x}^{2} & x > 1\end{matrix}\right.$

In this case, your limit actually exists.

${\lim}_{x \to 1} f \left(x\right) = 1$

However, $f \left(1\right) = \text{undefined}$.

In the first case, $f \left(a\right)$ is not defined and the limit does not exist. In the second case, $f \left(a\right)$ is not defined, but the limit exists.