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Sin X14) The inequality, 1 - 21 , holds when x is measured in radians and |x] < 1.sin X Find lim if it exists_ X-0...

Question

Sin X14) The inequality, 1 - 21 , holds when x is measured in radians and |x] < 1.sin X Find lim if it exists_ X-0

sin X 14) The inequality, 1 - 2 1 , holds when x is measured in radians and |x] < 1. sin X Find lim if it exists_ X-0



Answers

Find the limit (if it exists). Use a graphing utility to verify your result graphically.
$$\lim _{x \rightarrow \pi / 2} \frac{\sin x-1}{x}$$

For the given problem, we're going to evaluate the limit um As X goes to one And we have the sine of X -1. Yeah, Writing by two X -2. So with this in mind we're evaluating the limit as X goes for one. Which obviously if this was our function affects An act approaches one. If we look at f of one, we see that this would be an undefined value. However, if we're able to evaluate this As we get very close to one, such as 1.01 or 1.001, we see that it's extremely close to the value of .5 or one half. Similarly, if we look at this from the left, we see that as we get our actually this is going to be Your .99. Just the approach from left, we also get very close 2.5, so that's going to be our limit.

Okay, So for this problem, we want to find the limit as X approaches pi over two of one minus sine x sine X divided by coast X. So if we immediately just try substituting X equals pi by two in, we'll see that we would get one minus one, divided by zero, or we would get 0/0. So we get an indeterminant form. So it means that we have to try something else. It's a little bit, um, not quite obvious here, the way to proceed. But we only have a few different methods available to us, so we could try factoring something out. But we can see that we don't really have any factors, um, or anything to factor out. Or we could try rationalizing. But it's not immediately in a form where we can recognize that we can use the rationalization. But what we can do is use the Pythagorean identity that's sine squared. X plus co Squared X is equal to one, which means that's sine squared. X equals one minus coast squared X, or that sign X is equal to plus or minus the square root one minus coast squared X. Now what we can do here is substitute in this equation for sign. So I'm not going to be writing the limit over and over. I'm basically just going to be working on the expression here. The one minus sine X over co sex. We have one minus sine X over co sex sub making. That substitution will turn it into one minus the square root of one minus coast squared X over co sex for our purposes. Here, we're taking the We're only considering the positive route because we're in the region near X equals pi by two. So knowing the behavior of sign for coming to it from the left, then we'll be having pot. So actually all specify the left. Yeah, coming at it from the left, we'd use the positive route because Sine X is positive in that region to the left of pie by two. So starting off with looking at it from the left, we'd have one minus square root one minus coast squared X over coast X So what we can do now Multiply both sides by the conjugate of the numerator. So I have one plus the square root of one minus coast squared X divided by one plus route one minus coast squared X. So the numerator we would get one minus bracket one minus coast squared X. And in the denominator we'll have I won't multiply the co sex in for reasons that will become obvious. So we have Cossacks times one plus route one minus coast squared X in the denominator. Now in the numerator, we'd have one minus one plus coast squared X, and we'll have the denominator remaining unchanged for the moment so the numerator will become coast squared X. The denominator will become coast ax over one, plus route one minus coast squared X, which then means that we can fact, or we can cancel out or divide out one of the powers of coasts on top. So we'll get Cossacks on top and eliminate that Cossacks out front of the denominator. So I have Cossacks divided by one plus the square root of one minus coast squared X. Now, if we take the limit as X approaches pie by two from the left Here, the limit of one minus sine X over coast X by the rearranging that we just did is going to equal the limit as X approaches pie by two from the left of Cossacks over one, plus route one minus coast squared X where again, The fact that were coming in from the left is sort of encoded into the fact that we took the positive route for that sign X substitution. But with this expression, now we can evaluate this directly cause we'd have coast of pie by two on top and we have one plus route one minus coasts squared Hi by two. But actually, I just realized missed a chi made up here so that point there it's not pie by to that point where ah, sign is zero is pie. So that actually means that no matter which side were coming in from, uh, no. No matter how we are approaching pie by two, we'll be taking sign next to be positive. So, in fact, don't really need to specify or coming in from the left hand side of the right hand side. So we have coast of pie by two Costa pie. By two there is going to be zero and then we'd have division by one, plus the square root of one minus zero. So we get 0/2 or just equal to zero. Next, I'm just going to pop over to a graphing software and confirmed that. All right, so in Dez Mo's here I have The red line is the plot of one minus sine X over Cossacks in the blue line. This vertical line is a line at X equals pi by two. So if we zoom in very close, we can see that very certainly our function is going to be approaching zero as X approaches pie by two.

With this trigger metric problem, we can use direct substitution and substituting zero for X here. We know that sign of zero is zero. So that knocks out this whole first term and just leaves us with e minus one. So in the end, the final solution. So this limit as X approaches zero is negative one. And you could confirm that graphically as well. If you were to look at the graph of this sine function, it would also cross at negative one when X is zero.

Defined the limit as X approaches. Hi. Over Q of the equation. Sign next minus one. Oh, where squared sine X minus one. If we try to do this past substitution by substituting X into the equation, we find that sine X they would have positive one. And then the numerator we get zero on the denominator, we get zero. And because 0/0, which is undefined and that will not give us an answer Farley, we have to solve this in the week. One thing we can do is trying to multiply this equation by the con jacket of the denominator to try to cancel out. And he zeroes in the denominator. Do you have multiply by squared sine X plus one over. We're of signing less one. This will give us Kalim it as X approaches. Hi or two, uh, sine x minus. One times Aquarian Sign next, plus one over there sine x My, this one and says you can see we have a term on the numerator that is sine X minus one. We have a term in the denominator that is sine X minus one. They will cancel out, and this leads us. But the limit as X approaches. Pi over two, huh? The value sign. Sorry. Square root of sine X. Last one. We dropped that sign next. Minus swing terms. Now we can evaluate this box substitution. It will give us disk lewd sign. Hi over to plus one that isn't to the square root of one plus one which is recorded too.


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