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Re: Calculus Query
Quote:
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Originally Posted by Rickertsen2
this is kind of cheating but:
f(x)=0 where x=0
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Actually:
f(x)=0
f'(x)=0
f(x)^-1=1/0 which is a non-removable discontinuity. DNE
I dont think there is any, but then again I haven really taking time to look for a way.
Now you are gonna make me lose sleep on this. 
Dave
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