Problem

For the limit $$ \lim_{x \to 0}\frac{e^{2x} - 1…

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Problem 7 Medium Difficulty

For the limit
$$ \lim_{x \to 2}(x^3 - 3x + 4) = 6 $$
illustrate Definition 2 by finding values of $ \delta $ that correspond to $ \varepsilon = 0.2 $ and $ \varepsilon = 0.1 $.

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October 21, 2021

What does Illustrate Definition 2 mean?

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Video Transcript

Alright, let's have fun with limits. Alright, so here we have a limit as X approaches two of X cubed minus three X plus four and we can plug two in and we do get six. So we can feel pretty confident that limits correct. But it's kind of fun to investigate the whole idea of what does it mean to approach right? To kind of look at the more formal definition of the limits. So the idea is that if X is approaching two and getting really close, we can assume that we're within delta of two, so we can basically replace X By two plus a little delta because we're really close. And so each X can be replaced by to plus delta. And the goal is that that means we're approaching sex. Well we're that means we're getting close to six, right? So we'll say well we're close to six within kind of an epsilon. So the idea is that we kind of that if I'm getting close uh you know, to to for the X. Value that I should be getting close to six from my limits if the limits truly exists. So what we need to do now is clean up this thing and there's a lot of parts. So let's multiply it out. We can use the binomial theorem and pascal's triangle to know that the distribution for two plus sigma is too cute plus three times two squared times sigma plus three times to the first power times sigma squared plus one times sigma cubes. So hopefully you guys are pretty good with uh kind of a triple foil from pascal's triangle. And just a little review I pulled out these numbers, that's how this is like a one in front. The three The 3 and the one and then the to explain it kept reducing and the delta excellent kept increasing. So just little review. Okay then let's keep going. So we have still more terms but we have still we've only done the first term. Now we have to the middle term, I'm going to distribute the three. So I get minus six minus three delta plus four equal to six plus epsilon. Okay let's clean this up a little bit. So we get eight Plus, let's see 12. The delta Plus six Delta squared plus delta cubes -6 -3 Delta. Just cleaning up a little bit here and I'm gonna combine numbers and so on. Okay so let's see what we have here. We have a delta cubes. We have plus six delta square. So I took care of those two, I've got a 12 delta and a minus three delta. Is that all of my deltas there? Okay so that's a plus nine delta. And what do we have left? Eight minus six, that's 22 plus four is six. And um which means I can subtract six from both sides and I just get epsilon. Some not. Okay so um if I uh let's see here hold on a sec. Okay so we have our last expression now delta is very very small. And if I take a very small fraction and square it it gets even smaller and cubit smaller still. So if I were to assume that these higher order terms go to zero then I can say that a nine delta equals absence of not or delta is not expensive. Not. Well that is so funny. I added this up. Not to my epsilon because of physics. Ha ha permitted the free space. It's just good old epsilon. Not the primitive itty of free space anyway. Just epsilon. Okay, funny. Okay so delta is epsilon over nine? Okay so that means for different epsilon values. I can predict my delta. So let's see the two values that gave us um they gave us ε is .2. So then a delta would be .2 over nine which is point 0 to 19. So that is the delta associated with an epsilon to point to. And we've got one more to check. And that is if epsilon Is a .1 So then delta is .1 over nine And that is 0.0110. So um those are the two values that we're supposed to find associated with our epsilon. So anyway hopefully that helps you understand how we work on limits and why the formal definition of lim works. Okay, have a wonderful day. See you next time

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Problem

For the limit $$ \lim_{x \to 0}\frac{e^{2x} - 1…

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Kristen Karbon

Samuel Hannah

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Grace He

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals