How do you use the epsilon delta definition to find the limit of #x^3# as x approaches #0#?

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Answer 1 · 2018-05-04T10:29:43

Given any number #epsilon > 0# choose #delta_epsilon# such that:

#delta_epsilon < root(3)epsilon#

Then, for #x in (-delta_epsilon, delta_epsilon)# we have that:

#abs x < delta_epsilon#

#abs (x^3) = abs(x)^3 < delta_epsilon^3 < (root(3)epsilon)^3 = epsilon#

or shortly:

#x in (-delta_epsilon, delta_epsilon) => abs(x^3) < epsilon#

which proves the point.