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f(x)=1/ln(abs(x)) find the hole. I already found the V.A.:x=1,-1and H.A:y=0
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\[f(x)=1/\ln \left| x \right| \]
what is "the hole", what are V.A and what are H.A ?
the hole in the graph, Vertical asymptote, and Horizontal asymptote
he means when x=0
yes x=0.
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but how do you find that. I already now that is the answer.
*know
ln|x| is defined for 0<x<0, undefined for x=0, so you'll have a hole there
ln|x| , 1, are all continuous except at x=0. So 1/ln(abs(x)) is also cont. at all pts except x=0.
A hole is a removable singularity and the function does not have one.
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Thank you!!!
@oldrin.bataku the function does have it at x=0... right?
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