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Continuity of f(x)=1/logx 
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#1
May811, 05:43 AM

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Is the function f(x) = 1/logx discontinuous at x=0? My book says yes. It is continuous according to me. Can somebody verify?



#2
May811, 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}{\logx}<\epsilon[/tex] 


#3
May811, 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. 


#4
May811, 06:56 AM

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Continuity of f(x)=1/logx
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. 


#5
May811, 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. 


#6
May811, 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%2Flogx 


#7
May811, 10:55 AM

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


#8
May811, 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.



#9
May811, 05:31 PM

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The function f(x) = 1/logx for all x nonzero, 0 for x=0 is continuous.



#10
May811, 06:24 PM

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#11
May811, 10:48 PM

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Thanks to all!



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