# If there was a hole in the line at (2,3) and there is another point at (2,1), then would the graph be differentiable at that point and why?

Sep 5, 2016

If I understand the description correctly, the answer is no and the reason is below.

#### Explanation:

The short answer is that the function you have described is not continuous at $2$. It is a theorem that if $f$ is differentiable at $c$, then $f$ is continuous at $c$. Therefore non-continuous implies non-differentiable.

"A hole in the line at $\left(2 , 3\right)$" indicates to me that ${\lim}_{x \rightarrow 2} f \left(x\right) = 3$.

The point at $\left(2 , 1\right)$ implies that $f \left(2\right) = 1$

Now
$f ' \left(2\right) = {\lim}_{x \rightarrow 2} \frac{f \left(x\right) - f \left(2\right)}{x - 2}$

$= {\lim}_{x \rightarrow 2} \frac{f \left(x\right) - 1}{x - 2}$

This limit has the form $\frac{3 - 1}{2 - 2} = \frac{2}{0}$ which entails that the limit does not exist.

Since the derivative is the limit, the derivative also does not exist.