Question #c69a9

1 Answer
Andrea S. · Eddie
Apr 3, 2017

The limit:

#lim_(x->0) (1-3x)^3/x#

does not exist.

Explanation:

The numerator is continuous for #x = 0#, so we have:

#lim_(x->0^+) (1-3x)^3/x = lim_(x->0^+) (1-3x)^3 * lim_(x->0^+) 1/x = 1*+oo =+oo#

#lim_(x->0^-) (1-3x)^3/x = lim_(x->0^-) (1-3x)^3 * lim_(x->0^-) 1/x = 1*-oo =-oo#

Then we can see that the left and right limits are different which means that the function does not have a limit for #x->0#

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