Use the precise definition of a limit to prove the following l...

Question

Use the precise definition of a limit to prove the following limits. Specify a relationship between ε and δ that guarantees the limit exists.

lim x→0 x^2=0 (Hint: Use the identity √x2=|x|.)

Accepted Answer

Hi, everyone, let's take a look at this practice problem dealing with limits. This problem says to select the correct relationship between epsilon and Delta to prove that the limit as X approaches 0 of 5 X squad is equal to 0, using the epsilon-delta definition of a limit. We're given 4 possible choices as our answers. For choice A, we have Delta is equal to epsilon square divided by 5. For choice B, we have delta as equal to the square root of epsilon. For choice C we have Delta as equal to epsilon divided by 5. And for choice D, we have Delta as equal to the square root of the quantity of epsilon divided by 5. Now we're asked to prove that the limit as X approaches 0 of 5 x 2 is equal to 0, using the epsilon-delta definition of a limit. So, to use that definition, we first need to let epsilon be greater than 0, and then we need to find a delta greater than 0, such that, If 0 is less than the absolute value of X minus 0, And that's going to also be less than Delta. So, here, um, inside our absolute value, we have our variable X minus the value that X is approaching in our limit, which in this case is 0. So, if this statement is true, then the following statement must also be true, that the absolute value of our function in this case, which is 5 X squad, Minus the value that that limit evaluates to, which is, in this case, 0, that has to be less than epsilon. Now we can actually solve this inequality here for x. So first thing I noticed is that if I subtract 0 from 5 x2, I just get 5 X2, and taking the absolute value of that just gives me 5 x2, since 5 and X2 are both positive, so 5 x 2 is less than epsilon. Now I can solve this inequality for X by dividing both sides by 5, we'll get X squared. is less than epsilon divided by 5, then I'll take the square root of both sides. Then I'll get the absolute value of X is less than the square root of the quantity of epsilon divided by 5. And so, if I compare this expression with our statement involving X and Delta, we see that for both statements to be true, we need to choose delta to be equal to the square root of the quantity of epsilon divided by 5. And so this is the answer that I'm looking for, and if I come over here to my answer choices, the correct answer choice corresponds to answer choice D. So I hope that this explanation has been useful, and I'll see you in the next video.

Use the precise definition of a limit to prove the following limits. Specify a relationship between ε and δ that guarantees the limit exists.
lim x→0 x^2=0 (Hint: Use the identity √x2=|x|.)

Key Concepts

Limit Definition

Epsilon-Delta Relationship

Absolute Value and Squaring

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