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

Tags: continuity, fx1 or log|x|
 P: 1,395 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?
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 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 [itex]0
 HW Helper P: 6,164 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.
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## Continuity of f(x)=1/log|x|

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.
 PF Patron Sci Advisor Thanks Emeritus P: 38,424 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.
 P: 1,395 But Wolfram Alpha shows a continuous graph at x=0. http://www.wolframalpha.com/input/?i...9%3D1%2Flog|x|
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 No it does not. It shows a function where the two-sided limit at x=0 exists. That is not sufficient to establish continuity.
 PF Patron Sci Advisor Thanks Emeritus P: 38,424 Any numerical grapher can show only an "approximate" graph since it can calculate values only for a finite number of points.
 P: 429 The function f(x) = 1/log|x| for all x non-zero, 0 for x=0 is continuous.
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 Quote by 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.
I assume you mean x=1 for that function. It's well-defined at x=0, it's f(0)=1.
 P: 1,395 Thanks to all!

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