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

by zorro
Tags: continuity, fx1 or log|x|
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zorro
#1
May8-11, 05:43 AM
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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?
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arildno
#2
May8-11, 06:36 AM
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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]
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#3
May8-11, 06:53 AM
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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
#4
May8-11, 06:56 AM
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Continuity of f(x)=1/log|x|

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
#5
May8-11, 08:11 AM
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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.
zorro
#6
May8-11, 10:48 AM
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But Wolfram Alpha shows a continuous graph at x=0.
http://www.wolframalpha.com/input/?i...9%3D1%2Flog|x|
arildno
#7
May8-11, 10:55 AM
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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
#8
May8-11, 12:36 PM
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Any numerical grapher can show only an "approximate" graph since it can calculate values only for a finite number of points.
Jamma
#9
May8-11, 05:31 PM
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The function f(x) = 1/log|x| for all x non-zero, 0 for x=0 is continuous.
Char. Limit
#10
May8-11, 06:24 PM
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Quote Quote by HallsofIvy View Post
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.
I assume you mean x=1 for that function. It's well-defined at x=0, it's f(0)=1.
zorro
#11
May8-11, 10:48 PM
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Thanks to all!


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