Question
limx--->3- 5/x-3 I think it is -inf but I don't know how to prove it
Answers
Steve
if x<3, x-3 < 0
5/0 = ∞
but if the denominator is negative, it is -∞
Let u = x-3
Then you have the
limit (y -> 0-) 5/u
The limit is unbounded, and it is negative.
5/0 = ∞
but if the denominator is negative, it is -∞
Let u = x-3
Then you have the
limit (y -> 0-) 5/u
The limit is unbounded, and it is negative.