Question
I'm stumped on this one.
lim |x-1|sin(1/x-1)
x->1
lim |x-1|sin(1/x-1)
x->1
Answers
Steve
as written, there's no problem. Just plug in 1 to get
|1-1| (sin 1/1-1) = 0.
However, I think you mean
|x-1|sin(1/(x-1))
again, no problem. Since sin is always less than 1, we have 0 * z, where z<1, so zero again.
|1-1| (sin 1/1-1) = 0.
However, I think you mean
|x-1|sin(1/(x-1))
again, no problem. Since sin is always less than 1, we have 0 * z, where z<1, so zero again.