How do you prove limit of #root3(x)# as #x->0# using the precise definition of a limit?

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Answer 1 · 2017-03-08T14:10:02

For any #epsilon >0#, if #delta_epsilon < epsilon^3#

#abs (root(3)x - 0) < epsilon# for #x in (-delta_epsilon, delta_epsilon)#

For any value of #epsilon > 0# choose #delta_epsilon < epsilon^3#

Considering that:

#1) abs(root(3)x) = root(3)(abs x)#
#2) root(3)x# is strictly increasing for any #x#

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

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

so:

#abs (root(3)x - 0) < epsilon# for #x in (-delta_epsilon, delta_epsilon)#

which proves that:

#lim_(x->0) root(3)x = 0#