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

Question

Elijah Abram · Accepted Answer

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.

Christopher Allers · Answer

undefined To use the epsilon-delta definition to find the limit of x^3 as x approaches 0, we need to show that for any given epsilon (ε) greater than 0, there exists a delta (δ) greater than 0 such that if 0 < |x - 0| < δ, then |x^3 - 0| < ε. Let's proceed with the proof: Given ε > 0, we need to find a δ > 0 such that if 0 < |x - 0| < δ, then |x^3 - 0| < ε. |x^3 - 0| = |x^3| = |x|^3 Since we want to find a δ such that |x|^3 < ε, we can take the cube root of both sides: |x| < ε^(1/3) Now, we can choose δ = ε^(1/3). If 0 < |x - 0| < δ, then |x| < δ = ε^(1/3). Taking the cube of both sides, we have: |x|^3 < ε^(1/3)^3 = ε Therefore, if 0 < |x - 0| < δ = ε^(1/3), then |x^3 - 0| = |x|^3 < ε. Hence, we have shown that for any given ε > 0, there exists a δ > 0 such that if 0 < |x - 0| < δ, then |x^3 - 0| < ε. This satisfies the epsilon-delta definition of the limit. Therefore, the limit of x^3 as x approaches 0 is 0.