# How do you prove the statement lim as x approaches 0 for #x^3=0# using the epsilon and delta definition?

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To prove the statement lim as x approaches 0 for x^3=0 using the epsilon and delta definition, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if 0 < |x - 0| < delta, then |x^3 - 0| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a delta > 0 such that if 0 < |x - 0| < delta, then |x^3 - 0| < epsilon.

|x^3 - 0| = |x^3| = |x|^3

Since we want to prove that lim as x approaches 0 for x^3=0, we can assume that |x| < 1 (as x approaches 0, it gets arbitrarily close to 0).

Now, let's choose delta = epsilon^(1/3).

If 0 < |x - 0| < delta, then |x| < delta.

Since delta = epsilon^(1/3), we have |x| < epsilon^(1/3).

Taking the cube of both sides, we get |x|^3 < epsilon.

Therefore, if 0 < |x - 0| < delta, then |x^3 - 0| < epsilon.

Hence, we have proved that lim as x approaches 0 for x^3=0 using the epsilon and delta definition.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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