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 $L$?enter image description here

  • 2
    $\begingroup$ The limit still exists, it doesn't "care" about the specific value at the exact point. But the function will, obviously, not be continuous. $\endgroup$ – 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 $f(c)=L$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.