Question

Determine the points of discontinuity. State the type of discontinuity (removable, jump, infinite, or none of these) and whether the function is left- or right-continuous.
1. f(x) = lxl

2. g(t) = 1/(t^2-1)

3. f(x) = { cos (1/x) x does not = 0
{ 1 x = 0

4. f(x) = ln l x-4 l

Answer 1

absolute value is continuous

Answer 2

Thank you that is what I thought for the that one!

I also believe the answer to the second one is that there is an infinite discontinuity at t =1 and t=-1. But I am not positive.

Is the answer to the fourth that there is an infinite discontinuity when x<4 and it is right continuous?

The third one is giving me the most confusion.

Answer 3

yes for number 2

Answer 4

and number 4 since the log of zero is undefined, and it goes to \(-\infty\) there

Answer 5

for the third one, you have to compute the limit as \(x\to 0\) of \(\cos(\frac{1}{x})\)
if that limit exists, since cosine is continuous everywhere you can remove the discontinuity, or if that limit is 1, your function is continuous everywhere

Answer 6

unfortunately, as \(x\to 0^+\) we know \(\frac{1}{x}\to \infty\) and cosine will not have a limit, since it oscillated between -1 and 1 infinitely often
so there is not limit, and the discontinuity is not removable

Answer 7

however it not infinite limit, since it only goes between -1 and 1, not up to infinity
also it is not a jump, since the function is continuous everywhere except at 0