Question #bf3b9
2 Answers
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?
It might, if you specify it correctly.
If you had a vertical asymptote at
#lim_(x->a^(+)) 1/(x-a) = oo#
#lim_(x->a^(-)) 1/(x-a) = -oo#
However, if you choose the limit such that you approach
#lim_(x->a) 1/(x-a) = 1/0 = "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"
Alternatively, if you had a removable discontinuity, you might have a function
#f(x) = {(x, x < 1), (x^2, x > 1):}#
In this case, your limit actually exists.
#lim_(x->1) f(x) = 1#
However,
In the first case,