Is the function f(x) = 1/log|x| discontinuous at x=0? My book says yes. It is continuous according to me. Can somebody verify?
Note that all you have to show is that given some [itex]\epsilon>0[/itex], you can always find a [itex]\delta>0[/itex], so that for any x fulfilling [itex]0<x<\delta[/itex], we have: [tex]\frac{1}{|\log|x||}<\epsilon[/tex]
The function f is not defined for x=0 and this is a condition, so it is not continuous in 0. Note however, that if you expand the definition of f, so that f(0)=0, then it is continuous in 0.
My bad. Of course it is discontinuous at x=0, since it isn't defined there. I Like Serena has pointed out how to make a continuous extension of f, a feat that is possible since the limit of f at x=0 exists.
For another example, the function [tex]f(x)= \frac{x^2- 1}{x- 1}[/tex] is NOT continuous at x= 0 even though for all x except 0 it is equal to x+ 1 which is.
But Wolfram Alpha shows a continuous graph at x=0. http://www.wolframalpha.com/input/?i=f(x)=1/log|x|
No it does not. It shows a function where the two-sided limit at x=0 exists. That is not sufficient to establish continuity.
Any numerical grapher can show only an "approximate" graph since it can calculate values only for a finite number of points.