Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College To have a limit L as x approaches c, a function f must be defined on both sides of c and its values f(x) must approach L as x approaches c from either side ordinary limits are called two-sided. We're gonna look at one-sided limits. So if f fails to have a two-sided limit at c, it may still have a one-sided limit that is, a limit if the approach is only from one site. 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(x) equal L limit as x approaches c from the left of f(x) equal M if the two equal each other so if limit as x approaches c from the right of f(x) equals the limit as x approaches c from the left of f(x) then the limit as x approaches c of f(x) exists because L would equal M and hence the limit as x approaches c of f(x) would be L or M becuase they're the same thing A function as f(x) has a limit as x approaches c if and only if it has a left-hand and right-hand limits and these are equal so this means it can go either direction if we know that the limit of f(x) exisits, 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 gonna look at a theorem. We're gonna let limit as theta goes to zero of sine theta over theta equal one where theta's in radians We're gonna prove this, then we're gonna use it repeatedly um, in our homework and some of our examples So we're gonna start with drawing a picture we're gonna put a unit circle, so our OP is gonna 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 gonna equal AT over OA, OA is a distance of one so AT is really just gonna be tangent theta now we're also gonna restrict right now for theta to be in quadrant one which forces our zero less than theta less than pi halves. What that really is gonna do is it's gonna show all six of our trig functions being positive in the first quadrant and then we'll look at even and 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 OAP, O AP. The area this triangle would be less than the area of the sector of OAP. Remember a sector is a portion of a circle and that's gonna be less than the area of triangle OAT. OAT when we look at this we know that the area to triangles one-half base height we know that the area a sector is one-half theta r squared if you've forgotten that think about a whole circle area is pi r squared and we don't want the full circle we only want a portion of it let's call the portion theta out of two pi or theta out of a full circle the pis then cancel and we get one-half theta r squared making sure were in radians the area of the triangle is one-half base height so when we look at OAP. OAP one-half the base here because it's a unit circle is one the height is sine theta when we look at the sector we have one-half theta radius squared well it's a unit circle so the radius is one and when we look at triangle OAT one-half base height, we know that our base is one again 'cause it's a unit circle and we had determined a moment ago that our height this just tangent theta so one-half one times tangent theta if I multiplied everything through by a two to get rid of the one halfs, we get sine theta less than theta less than tangent theta now we're gonna divide everything through by sine theta if I divide by sine theta or multiply by one over sine theta the sine theta cancels, gives me one less than theta over sine theta less than tangent's just sine over cosine and I'm gonna multiply by one ever sine so the sines cancel leaving the one over cosine now we're gonna take the reciprocal function of everything. When we take the reciprocal function, we have to flip our inequality signs. So the reciprocal of one is just one. one over one 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 gonna use that sandwich theorem or that squeeze theorem and if I can show that this cosine theta is gonna go to one and this is one, then I know the function in between would also have to go to one so what happens to limit as theta approaches zero from the right hand side zero from the right hand side of cosine, cosine a zero is just one so we know that the left and the right both are going to go to one as theta goes to zero 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(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 zero 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 right have to be the same via the symmetry. Therefore, the limit of theta going to zero 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 zero also we have limit laws where the limit laws as x as x approaches positive and negative infinity limit as x goes to positive negative infinity of f(x) equals L and limit as x goes to positive negative infinity of g(x) equals 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 gonna also show that the limit as h goes to zero of cosine h minus one over h is zero the proof of this is start out with the cosine h minus one over h and multiply by its conjugate so when we foil this out, we get cosine squared h minus one over h times cosine h plus one now we know cosine squared h minus one is really negative sine squared h by the pythagorean identity because of the limit laws, I can now separate things up. So I'm gonna have sine h over h times negative sine h over cosine h plus one once again, from our limit laws, I can separate these up into separate limits that are then multiplied. Limit as h goes to zero of sine h over h times the limit as h goes to zero of negative sine h over cosine h plus one we know that the limit as h goes to zero of sine h over h is just one 'cause we proved just a moment ago this one, we're just gonna literally stick zero in for h. What is sine of zero? it's zero. What is cosine of zero? It's one we could actually officially make that a negative if we want so zero over anything is zero, zero times one is zero. This one's gonna be very important also if we look at a picture and some exercises, were gonna have the limit as negative one goes to the right negative one from the right, so as we get closer and closer and closer from this side, what happening to our y? our y-value is going to one. So this is a true statement if we look at x going to zero from the left what happening to our y? Our y is going to zero, so this one's a false statement limit as x goes to zero of f(x) exists. All that's asking is is the left side and the right side going to the same values. They're both going to zero so that would be true if we look at limit as x goes to zero of f(x) equals one that's false because it's going to zero, 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(x), so when we look at one coming in from the left side would 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 zero because these two are not equal this does not exist so it's gotta be false negative limit as x goes to negative one from the left well there aren't any values here so this one does not exist 'cause there is no y to look at so that's true there 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 zero of sine 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 gonna do is we're actually gonna multiply by k to get the kt and the kt here to be the same but I can't just throw a k in the bottom without keeping it equivalent, so I'mgoing to throw a k in the top when I do this, now I can think about sign of kt over kt times the limit as t goes to zero of k well this, by our identity that we proved a few minutes ago is one and the limit as t goes to zero of any constant is just a constant so the final answer there would just be k looking at another example we have a sine of three h in the denominator, but we only have h in the numerator so I'm gonna multiply by three, so I get three h and three h but to keep it balanced, if I put a three up top, I gotta put a three in bottom now when we look at this we know that this first part is gonna 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 as h goes to zero of a constant is just really the constant so the limit there is one-third we can bring in other trig functions, so if I have two t, I 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 the only thing we really know at this moment is that t over sine t as t goes to zero is really one so we're gonna pull that out and we're gonna leave everything else as another separate function so now we get the limit as t goes to zero of t over sine t is one. If I put in zero for cosine t cosine of zero is one. So two times one 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 gonna do is I'm gonna split everything up into sines and cosines. Cotangent is cosine over sine cosecant is one over sine to make this work when I see this sine x I need to have an x on top when I see a sine two x, I need a two x on top so if I put in an- and then I'm gonna put everything else down here at the end for a second so if I put in an x on top, to keep the equation balanced we have to put an x in the bottom if I put it two x on top, to keep the equation balanced, I've gotta put a two x in the bottom literally I'm just multiplying by things so that I can get them to have pieces that will equal one so this x over sine x is gonna be one. Two x over sine two x is gonna be one and then we need to evaluate whatever this last piece is gonna turn into so we're gonna get limit as x goes to zero of x over sine x times the limit as x goes to zero and two x over sine two x times the limit as x goes to zero and this is gonna simplify up to three cosine x so when we evaluate this we're gonna get one times one times three, the cosine of zero is one so this whole thing is gonna just end up being equal to three looking at another one here we're gonna do some substitution. If we thought about let sine of h equal some theta then when sine h goes to zero as h goes to zero so if sine h goes to zero as h goes to zero and we know that the limit as sine h goes to zero as h goes to zero, theta would also have to then go to zero, 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 and another one. Limit y approaches zero, sine three y I'm gonna leave on top with nothing in the bottom for a second cotangent is gonna be cosine over sine I'm gonna leave the sine of five y by itself for a moment the y I'm gonna put way down here at the end and cotangent, but cotangent's in the denominator now, so we're gonna have sine of four y in the top and cosine four 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 a means I'm gonna put a three y on the top I'm gonna put a five y on the top which means I'm gonna put a five y in the bottom I need a four y in the bottom, so I'm gonna put a four y on the top when we look at this and break it into pieces we get the limit as y goes to zero of sine three y over y times the limit as y goes to zero of five y over sine five y times the limit as y goes to zero of sine four y over four y times the limit as y goes to zero if we simplify this up, this y and this y, that y and that y are gonna cancel, so we're gonna get 12 cosine five y over cosine four y so now we know we have one times one times one times when I stick in zero, 5 times zero's zero but the cosine of zero is one. 4 times zero's zero the cosign of zero's one, so we're gonna get 12 times one over one or a final answer of 12 oh I lost this five, whoops. How about this five there too so twelve-fifths I lost that five that five there okay, so twelve-fifths our last example is gonna be a little different and we're gonna pull out an x and get one plus cosine x over sine x cosine x so now we're gonna rewrite this as x sine x times limit as x goes to zero of one plus cosine x over cosine x one times one plus one over one or just a final answer of two thank you and have a wonderful day