Does the limit exist when there is a removable discontinuity but the function takes a different value there?

I know that when there is a hole the limit still exists because the limit is asking what value the function approaches, but does the limit still exist at the hole when the function takes a different value? In the graph on the right is the limit at $x=c$ equal to $L$?

• The limit still exists, it doesn't "care" about the specific value at the exact point. But the function will, obviously, not be continuous. – Clement C. Mar 24 '17 at 15:16

Yes, the limit still exists and it has the same value, so it is still $L$ and not $f(c)$.
It's important to understand that the limit of a function $f$ at a point $c$ (its existence and its value if it exists) is completely determined by the function values of $f$ near $c$ but not at $c$.
Intuitively you could say the limit tries to "predict" $f(c)$ based on the neighboring values of $f$ and it predicts a real number $L$ if the function can be made continuous by setting $f(c)=L$.