How do you find the limit of #(x^3 *abs(2x-6))/ (x-3)# as x approaches 3?

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Answer 1 · 2016-10-02T15:59:12

The one sided limits disagree, so there is no two sided limit.

Note that if #t >= 0# then #abs(t) = t# and if #t <= 0# then #abs(t) = -t#.

When #x > 3# then #x - 3 > 0# and we have:

#(x^3*abs(2x-6))/(x-3) = x^3*abs(2(x-3))/(x-3) = (x^3*2(color(red)(cancel(color(black)(x-3)))))/(color(red)(cancel(color(black)(x-3)))) = 2x^3#

Hence #lim_(x->3^+) (x^3*abs(2x-6))/(x-3) = lim_(x->3^+) 2x^3 = 54#

When #x < 3# then #x - 3 < 0# and we have:

#(x^3*abs(2x-6))/(x-3) = x^3*abs(2(x-3))/(x-3) = (x^3*(-2(color(red)(cancel(color(black)(x-3))))))/(color(red)(cancel(color(black)(x-3)))) = -2x^3#

Hence #lim_(x->3^-) (x^3*abs(2x-6))/(x-3) = lim_(x->3^-) -2x^3 = -54#

Since the one sided limits disagree there is no two sided limit as #x->3#.