# Continuity of f(x)=1/log|x|

1. May 8, 2011

### 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?

2. May 8, 2011

### 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:
$$\frac{1}{|\log|x||}<\epsilon$$

3. May 8, 2011

### 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.

4. May 8, 2011

### arildno

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.

5. May 8, 2011

### HallsofIvy

Staff Emeritus
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.

6. May 8, 2011

7. May 8, 2011

### 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.

8. May 8, 2011

### HallsofIvy

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

9. May 8, 2011

### Jamma

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

10. May 8, 2011

### Char. Limit

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

11. May 8, 2011

### zorro

Thanks to all!