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Posted

Possibly neither. You cannot plot all the infinite pairs (x,y=f(x)) in an interval but only a few. If the program just draws the function value at some roughly equidistant points x_i, then there's a chance that none of them equals 0.5. Btw.: There is no reason to assume a spike shape. Your function is one everywhere except at x=0.5 where it is not defined -> no non-zero slope anywhere.

 

Edit: Corrected x=0 to x=0.5

Posted (edited)

It's even worse:

http://www.wolframalpha.com/input/?i=1%2Fx+for+x%3D0

http://www.wolframalpha.com/input/?i=-1%2Fx+for+x%3D0

 

To explain why I consider that even worse: 1 is the correct limit for 1^(1/x) when x goes to zero. So the function is not defined there but at least the limit exists. In contrast, the limit of 1/x when x->0 does not exist and can as well be the negative of the value claimed by alpha when zero is approached from the left, i.e. when approaching zero from the negative values.

 

Funny that my links get cast into an [ url] ... [ /url] box automatically.

 

EDIT: Clarified "from the left".

Edited by timo

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