Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. To have a limit L as X approaches CA function, F must be defined on both sides of C and its value F of X must approach L as X approaches C from either side. Ordinary limits are called 2 sided. We're going to look at one sided limits. So if F fails to have a 2 sided limit at C, it may still have A1 sided limit. That is a limit if the approach is only from one side. If the approach is from the right, the limit is a right hand limit, and if it's from the left, a left hand limit. So limit as X approaches C from the right of F of X equal L limit as X approaches C from the left of F of X equal M if the two equal each other. So if limit as X approaches C from the right of F of X equals the limit as X approaches C from the left of F of X, then the limit as X approaches C of F of X exists because L would equal M and hence the limit as X approaches C of F of X would be L or M because they're the same thing. A function F of X has a limit as X approaches C if and only if it has a left hand and right hand limit, and these are equal. So this means it can go either direction. If we know that the limit of F of X exists, we know that it has both a left side limit and a right side limit. If we know it has a left side limit and a right side limit, and the left and the right limits are the same, then we know the original limit of the function at that location has to exist. We're going to look at a theorem. We're going to let limit as Theta goes to 0 of sine Theta over Theta equal 1, where Theta is in radians. We're going to prove this and we're going to use it repeatedly in our homework and some of our examples. So we're going to start with drawing a picture. We're going to put a unit circle, so our op is going to be one. We're going to draw any old point on the unit circle. So if I drop down my perpendicular, I know that OQ is cosine Theta and I know that PQ is sine Theta. So we know that by similar triangles, the sine Theta divided by cosine Theta is going to equal AT over OA. OA is a distance of 1, so AT is really just going to be tangent Theta. Now we're also going to restrict right now for Theta to be in quadrant one, which forces our 0 less than Theta less than π halves. What that really is going to do is it's going to show all six of our trig functions being positive in the first quadrant and then we'll look at even an odd functions to determine what happens in the other quadrants. So looking at this triangle, actually multiple triangles, if we looked at the area of triangle OAPOAP, the area of this triangle would be less than the area of the sector of OAP. Remember this sector is a portion of a circle and that's going to be less than the area of triangle OATOAT. When we look at this, we know that the area of a triangle is 1/2 base height. We know that the area of a sector is 1/2 Theta r ^2. If you've forgotten that, think about a whole circle areas Pi R-squared. And we don't want the whole circle, we only want a portion of it. Let's call that portion Theta out of 2π or Theta out of a full circle. The pi's then cancel and we get 1/2 Theta R-squared, making sure we're in radiance. The area of the triangle is 1/2 base height. So when we look at OAPOAP 1/2 the base here because it's a unit circle is 1. The height is sine Theta. When we look at the sector we have 1/2 Theta radius squared. Well, it's a unit circle, so the radius is 1. And when we look at triangle OAT 1/2 base height, we know that our base is one again because it's a unit circle. And we had determined a moment ago that our height is just tangent Theta. So one half one times tangent Theta. If I multiplied everything through by A2 to get rid of the one halfs, we'd get sine Theta less than Theta less than tangent Theta. Now we're going to divide everything through by sine Theta. If I divide by sine Theta or multiply by one over sine Theta, the sine Theta cancel gives me one less than Theta over sine Theta, less than tangents, just sine over cosine, and I'm going to multiply by one over sine. So the Sine's cancel leaving me one over cosine. Now we're going to take the reciprocal function of everything. When we take the reciprocal function, we have to flip our inequality sign. So the reciprocal of one is just one 1 / 1 reciprocal of Theta over sine Theta is sine Theta over Theta and the reciprocal of one over cosine Theta is cosine Theta. Now we're going to use that sandwich theorem or squeeze theorem. And if I can show that this cosine Theta is going to go to 1 and this is 1, then I know the function in between would also have to go to 1. So what happens to limit as Theta approaches zero from the right hand side? Zero from the right hand side of cosine. Cosine of 0 is just one. So we know that the left and the right both are going to go to one as Theta goes to 0. So we know that the sine Theta over Theta has to also go to one. Now we've currently only proven this. Going zero from the right hand side, we know that sine Theta and Theta are both odd functions. Therefore F of Theta equaling sine Theta over Theta is an even function. So we have symmetry about the Y axis. We know that limit of Theta of 0 from the right side of sine Theta over Theta. Because of the symmetry about the Y axis, we know that from the left side it also has to equal the same value because the Y values on the left and the right have to be the same by the symmetry. Therefore the limit of Theta going to 0 of sine Theta over Theta is one. We also want to pay attention to this reciprocal function here because Theta over sine Theta equals one as Theta goes to 0. Also we have limit laws where the limit laws as X as X approaches positive or negative Infinity, limit as X goes to positive or negative Infinity of F of X equal L and limit as X goes to positive negative Infinity of G of X equal M Then we can add the two, we can subtract the two. We had the product, the constant, the quotient, the power, the same as before, but now X is going to positive and negative Infinity. We're going to also show that the limit as H goes to 0 of cosine H - 1 / H is 0. The proof for this is start out with the cosine H - 1 / H and multiply by its conjugate. So when we foil this out, we get cosine squared H - 1 / H times cosine H + 1. Now we know cosine squared H -, 1 is really negative sine squared H by the Pythagorean identity. Because of the limit laws, I can now separate things up. So I'm going to have sine H / H times negative sine H over cosine H + 1. Once again, from our limit laws, I can separate these up into separate limits that are then multiplied. Limit as H goes to 0 of sine H / H times the limit as H goes to 0 of negative sine H over cosine H + 1. We know that the limit is H goes to 0. Sine H / H is just one because we proved it a moment ago. This one we're just going to literally stick 0 in for H. What is sine of 0? It's zero. What is cosine of 0? It's one. We could actually officially make that a negative if we want. So zero over anything is 00 times one is 0. This one's going to be very important. Also, if we look at a picture and some exercises, we're going to have the limit as -1 goes to the right -1 from the right. So as we get closer and closer and closer this direction, what's happening to our Y? Our Y value is going to 1. So this is a true statement. If we look at X going to zero from the left, what's happening to our Y? Our Y is going to 0. So this one's a false statement. Limit as X goes to 0 of F of X exists. All that's asking is the left side and the right side going to the same values. They're both going to 0. So that would be true if we look at limit as X goes to 0F of X equal 1. That's false because it's going to 0. The Y's are approaching not what it actually equals at that location, but what the Y's are approaching. X approaches one of F of X. So when we look at one coming in from the left side, we'd get the limit as X goes to one from the left. the Y values are going to one, but the limit as X approaches one from the right. If we're looking at it this way, the Y values are going to 0 because these two are not equal. This does not exist. So it's got to be false negative limit as X goes to -1 from the left. Well, there aren't any values here, so this one does not exist because there is no Y to look at. So that's true. There are a few examples of reading information off of the graph. Let's look at some more examples here. If we have limit as T goes to 0, assign KT over T. We know that we have a property that says if this angle here was Theta and it was Theta down here, then that would have to go to one. Well, it's currently not the same angle measures. So what we're going to do is we're actually going to multiply by K to get the KT and the KT here to be the same. But I can't just throw AK in the bottom without keeping it equivalent. So I'm going to throw AK in the top. When I do this now I can think about sine of KT over KT times the limit as T goes to 0 of K. Well this by our identity that we proved a few minutes ago is 1 and the limit as T goes to 0 of any constant is just a constant. So the final answer there would just be K. Looking at another example, we have a sign of three H in the denominator, but we only have H in the numerator. So I'm going to multiply by three. So I get 3H and 3H. But to keep it balanced, if I put a three in top, I've got to put a three in bottom. Now when we look at this, we know that this first part is going to go to one. We could think of writing it as two separate limits that are then multiplied together. When we do this, we know that this first one goes to one. The second one the limit is H goes to zero of a constant is just really the constant, so the limit there is 1/3. We can bring in other trig functions. So if I have two TI, know that tangent is sine divided by cosine. So I could think of this as the limit as T approaches zero of two T cosine T over sine T. But the only thing we really know at this moment is that T over sine T as T goes to 0 is really one. So we're going to pull that out and we're going to leave everything else as another separate function. So now we get the limit as T goes to 0 of T over sine, T is one. If I put in zero for cosine T, cosine of 0 is 1. So 2 * 1 is one. So the limit here is just two. Now they can get much more complex. Let's look at this one. The first thing I'm going to do is I'm going to split everything up into sines and cosines cotangent as cosine over sign. Cosecant is one over sign. To make this work, when I see the sign XI need to have an X on top. When I see a sign 2 XI need a 2X on top. So if I put it and then I'm going to put everything else down here at the end for a second. So if I put an X on top, to keep the equation balanced, we have to put an X on the bottom. If I put a 2X on top to keep the equation balanced, I've got to put a 2X on the bottom. Literally, I'm just multiplying by things so that I can get them to have pieces that will equal 1. So this X over sine X is going to be 12X over sine 2X is going to be 1. And then we need to evaluate whatever this last piece is going to turn into. So we're going to get limit as X goes to 0 of X over sine X times the limit as X goes to 0 of 2X over sine 2X times the limit as X goes to 0. And this is going to simplify up to three cosine X. So when we evaluate this, we're going to get 1 * 1 * 3. The cosine of 0 is 1. So this whole thing is going to just end up being equal to three. Looking at another one here, we're going to do some substitution. If we thought about let sine H equals some Theta then when sine H goes to 0 as H goes to 0. So if sine H goes to 0 as H goes to 0, then we know that the limit as sine H goes to 0 as H goes to 0. Theta would also have to then go to 0 because sine H is equal to Theta, and we'd get sine of Theta over Theta, which would just equal one. That one's a little tricky, but think of this whole thing as some angle, and this is the same angle, and hence then it's got to equal the same thing, which is the one looking at another one. Limit Y approaches 0 sine 3 Y I'm going to leave on top with nothing in the bottom for a second. Cotangent is going to be cosine over sine. I'm going to leave the sine of five Y by itself for a moment. the Y I'm going to put way down here at the end and cotangent, but cotangents in the denominator now. So we're going to have sine of four Y in the top and cosine 4 Y in the bottom. When I look at this, in order for this equation to work, I have to put a three Y in the bottom here, which means I'm going to put a three Y in the top. I'm going to put A5 Y in the top, which means I'm going to put A5 Y in the bottom. I need a four Y in the bottom, so I'm going to put a four Y in the top. When we look at this and break it into pieces, we get the limit as Y goes to 0, a sine three y / y times the limit as Y goes to 0A5Y over sine, 5 Y times the limit as Y goes to 0. Assign four y / 4 Y times the limit as Y goes to 0. We simplify this up, this Y and this Y, that Y and that Y are going to cancel. So we're going to get 12 cosine 5Y over cosine 4Y. So now we know we have 1 * 1 * 1 times when I stick in 05 times 00, but the cosine of 0 is 1/4 times 00. The cosine of 0 is 1. So we're going to get 12 * 1 / 1, or a final answer of 12. Oh, I lost this 5. Oops. How about A5 there too? So 12 fifths. I lost that five, that five there. OK, so 12 fifths. Our last example is going to be a little different, and we're going to pull out an X and get one plus cosine X over sine X cosine X. So now we're going to rewrite this as X sine X times the limit as X goes to zero of one plus cosine X over cosine X 1 * 1 + 1 / 1 or just a final answer of 2. Thank you and have a wonderful day.