How do you evaluate # (x)/((x^2)-x)# as x approaches 0?

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How do you evaluate # (x)/((x^2)-x)# as x approaches 0?

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Answer 1 · 2016-04-05T20:10:40

#lim_(x->0) x/(x^2-x) = -1#

Explanation:

In the given form, both the numerator and denominator are zero when #x=0#, but when #x != 0# we can divide both numerator and denominator by #x# to find:

#lim_(x->0) x/(x^2-x) = lim_(x->0)1/(x-1) = 1/(0-1) = -1#

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How do you evaluate # (x)/((x^2)-x)# as x approaches 0?

1 Answer

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