Is f(x) = 1/log|x| Continuous at x=0?

[zorro]

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?

 

[arildno]

Note that all you have to show is that given some $\epsilon>0$, you can always find a $\delta>0$, so that for any x fulfilling $0<x<\delta$, we have:
$$\frac{1}{|\log|x||}<\epsilon$$

 

[I like Serena]

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.

 

[arildno]

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.

 

[HallsofIvy]

For another example, the function
$$f(x)= \frac{x^2- 1}{x- 1}$$
is NOT continuous at x= 0 even though for all x except 0 it is equal to x+ 1 which is.

 

[zorro]

But Wolfram Alpha shows a continuous graph at x=0.
http://www.wolframalpha.com/input/?i=f(x)=1/log|x|

 

[arildno]

No it does not.
It shows a function where the two-sided limit at x=0 exists.
That is not sufficient to establish continuity.

 

[HallsofIvy]

Any numerical grapher can show only an "approximate" graph since it can calculate values only for a finite number of points.

 

[Jamma]

The function f(x) = 1/log|x| for all x non-zero, 0 for x=0 is continuous.

 

[Char. Limit]

For another example, the function
$$f(x)= \frac{x^2- 1}{x- 1}$$
is NOT continuous at x= 0 even though for all x except 0 it is equal to x+ 1 which is.

I assume you mean x=1 for that function. It's well-defined at x=0, it's f(0)=1.

 

[zorro]

Thanks to all!