Enroll in one of our FREE online STEM summer camps. Space is limited so join now!View Summer Courses

University of North Texas

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Problem 71
Problem 72
Problem 73
Problem 74
Problem 75
Problem 76
Problem 77
Problem 78
Problem 79
Problem 80
Problem 81
Problem 82
Problem 83
Problem 84
Problem 85
Problem 86
Problem 87
Problem 88
Problem 89
Problem 90
Problem 91
Problem 92

Need more help? Fill out this quick form to get professional live tutoring.

Get live tutoring
Problem 88

For what values of $ a $ and $ b $ is the following equation true?

$$ \displaystyle \lim_{x\to 0} \left( \frac{\sin 2^x}{x^3} + a + \frac{b}{x^2} \right) = 0 $$

Answer

Check back soon!

You must be logged in to like a video.

You must be logged in to bookmark a video.

## Discussion

## Video Transcript

we want to find values of A and B for the following limit as experts zero. That makes it equal to zero. So if we want this to be true, well, first, let's go ahead and just simplify this a little bit. So the first thing we can do is distributed across the pluses, and really, I'll just go ahead and group it like this to start. So the limit as X approaches zero, uh, sign of two x over execute. And actually, I'll go ahead and combined these two just by adding them really quickly. So in the new mirror, that's gonna be a plus B X all over, and then we'll have the x cute. So we'll have that and will have plus the limit as ex approach zero of a 0 to 0. And we know that a is a constant with respect to X. So this year will just end up being a so we can go ahead and some truck that over and so will get negative. A is equal to the limit as X approaches zero of signed to X plus bx over x cute. So we have this now. Now, if we were to just plug this in directly. Well, we're going to have zero. They're zero. They're enduring the denominator. So we have zero over zero, which means we can imply low petals rule to this. So for this next up, we're going to apply Loki tolls. And so that's going to give us. That's the limit. As X approaches zero of what the derivative of sign is co sign to X, and then we have to apply chain rules that we take the drone on the inside. So you multiply that by two, and then we take the derivative of B X, which just leaves us with B and then in the denominator will have three x squared. And this year should still be equal to negative, eh? Okay, Now, if we were to apply the limit directly, so co sign of zero is one. So, Wendell, with two plus b and then the denominator, we end up with zero. So we still can't divide by zero. But we want this limit toe hold. So that means So this implies we want our numerator toe also be zero. So we're gonna want to plus B to be equal to zero, which implies B is equal to negative too. So we found one value. So I'm just gonna go ahead and put that up here at the start. So B is equal to two, and now we can go ahead and plug that in for B. And doing that will give us something that we can take the derivative with respect to Earth. Apply low petals rule. Since, what? Zero over zero. So I'll just go ahead and apply Loki tall to walk in. So that's going to be the limit as X approaches zero of So we're going to have too. So the derivative of co sign is going to be negative. Sign to X, and then we have that, too, on inside that we have to take the derivative look that out front. So this too would actually become a four. And then we divide this by so we'd get X. And then we have to move that too out front by power room. And so then we'd get two times three or six. Okay. No, If we were to apply the lead here again, Sign of zero is gonna be zero X is just going to be zero. So we can imply. Loki told you one more time and I'll go ahead and move this up here. So by Loki tolls rule again. And actually, one thing I should still have over here is that the right hand side of our equation should still be equal to negative eight. Yeah. So now, using low petals ruled, the left hand side is going to become the limit as Ex purchase zero. Oh, so we take the derivative of the numerator again. So the derivative of sign is going to be co side to X. Then we have to take the derivative of change rule. So we end up with another two, multiply that by the negative for we already had that becomes negative eight and then in the denominator, the derivative of six X, which is be six. And this here is still equal to negative. Now we can go ahead and applied the limit again and again. Co sign of zero is going to be one. So we'll end up with negative eight over six. Busy, eh? Are negative, eh? So we could multiply the negative over and simplify it over six to give us 4/3. So we end up with that. A is equal to 4/3 and B is equal to negative too.

## Recommended Questions