episode 5 more techniques for computing limits so last time we looked at evaluating some limits some of them were pretty easy if we would just try to plug in x equals a into f of X and if we get a number like we do with a polynomial or a rational function where the bottom is not zero that's just gonna be our limit right there they're pretty easy to evaluate sometimes we didn't really talk about this last time but if we're at the edge of a domain or we're two pieces of the piecewise function match up then we may not actually that that number that we get may not actually be the limit but those are pretty rare sometimes we don't get a number though sometimes we have a number divided by zero that gives us some more work to do if that was a nonzero number divided by zero then we'd have our infinite limit situation would be positive infinity negative infinity or the to one side that's what disagree the last time we started looking at what to do when we had zero over zero limits that was very different than non zero over zero the tool that we used was the the idea that if two functions agree everywhere except for a then their limits would be the same as X approaches a the idea here was that we looked at rational functions that were zero over zero and we realized that we could factor them and cancel the problem so let's try some more examples of that so the limit as X approaches 3 of x cubed minus 5 x squared plus 3x plus 9 divided by X minus 3 when I plug in 3 I get 27 minus 45 plus 9 plus 9 that's 0 and then of course 3 minus 3 is 0 so the idea still holds I need to factor the top and 1 of I need to factor the top and cancel with the X minus 3 in the bottom but that's a cubic polynomial cubic polynomials you're a lot harder to factor but we got one thing working in our advantage I know that X minus 3 has to be one of the factors I know that because I plugged in 3 got zero and I also need it to cancel with the X minus three in the bottom right that's the only way I'm get rid of my zero for zero problem so we'll just use long division right so if we divide X minus 3 into X cubed minus 5 x squared plus 3x plus 9 I need to figure out what I need to multiply X by to get X cubed that would be x squared so x squared times X minus 3 is X cubed minus 3x squared subtract that X minus 3s cancel as I set them up to do and I end up with minus 2 x squared then I bring down the 3x so what do I multiply X by to get minus 2 x squared well that would be minus 2 X so then I get I also have a plus 6 X and so what's left over is minus 3x what do I need to multiply X by to get minus 3 X and that's minus 3 so multiply there I get minus 3x plus 9 hey that's the thing I exactly wanted so I don't have a remainder so I know that my cubic polynomial X cubed minus 5 x squared plus 3x plus 9 I know that that's equal to X minus 3 times x squared minus 2x minus 3 hey so I've got it factored to the point that I needed to be factored right so I can cancel the X minus 3 there's my 0 over 0 problem now gone away so I'm left with the limit as X approaches 3 of x squared minus 2x minus 3 I can plug that in without any trouble alright I get 9 minus 6 minus 3 and so this limit is 0 0 is a number I'm not dividing by 0 here so this limit is 0 so like I said last time I don't really need you to rationalize the denominator and your answer but I do expect you to know how to do it the idea here is that we want to make the bottom part of the fraction be a ration it'll probably be an integer so here I have four divided by two minus the square root of seven to two minus the square root of seven is irrational so this number has an irrational denominator so what we do is multiply by the conjugate of that to make it a rational number on the bottom okay the conjugate is just changing that minus sign to a plus or possibly a plus to a minus sign but we need it here it'll always be a minus sign changing to a plus all right so we multiply both the top and the bottom by two plus the square root of seven we're multiplying by one so we haven't changed anything but look at what it does to the bottom right when we foil out the bottom we get 2 times 2 is 4 2 times square root of 7 minus 2 times square root of 7 hey those will cancel and then minus square root of 7 times plus square root of 7 is minus 7 so the bottom is going to be negative 3 hey that's that's an integer value so this is minus 4 times 2 plus the square root of 7 divided by 3 equal in value to the thing that we found earlier except the denominator is rational the reasoning behind this is kind of interesting it goes back to the days before calculators when you would get a decimal approximation of this number by walking over to the shelf and pulling off a book and looking up the value of the square root of 7 and then by hand do the division and it's so much easier to divide by 3 by hand than it is by dividing by that 2 minus the square root of 7 so that's why we prefer it's not such a big deal now that we can easily find a calculator but the idea of doing this can be very very helpful when we have a 0 over 0 limit that's caused by a subtraction involving a square root an example like this the limit as X approaches 0 of the square root of 1 plus X minus the square root of 1 - x / x right if I try to plug in 0 I get 1 minus 1 divided by 0 so this is 0 / 0 and we can't factor it like we did some of the other examples that we did because the top is not a polynomial but if we try to rationalize the numerator that is getting rid of the square roots in the numerator with subtraction we can get around this problem it'll put us in a position where we can cancel the 0 / 0 problem right so the conjugate here would be the square root of 1 plus X plus the square root of 1 minus X right if we are careful about the way we foil that out right the first times the first would just give us 1 plus the square root of x the the outer terms would give us the square root of 1 plus x times 1 minus X the product of the inside terms would be minus the square root of 1 minus x times 1 plus X and then the product of the last term's would be minus parentheses 1 minus X so look closely that the two things in the middle the outside of the inside terms will cancel and so the top we also have a 1 that cancels and so what we end up in the top with it is is just 2x and then the bottom there is no reason to distribute that X right at the bottom is just x times the square root of 1 plus X plus the square root of 1 minus X hey there's my 0 / 0 problem right X divided by X so those will cancel and so this is the limit as X approaches 0 of 2 divided by the square root of 1 plus X plus the square root of 1 minus X and if I try to plug 0 into this I don't run into any problems right the bottom is now and so we didn't have two things that do not cancel each other out right so we end up with two divided by one plus one that's equal to one so this limit is one so we had a function that had a subtraction involving two square roots and we rewrote that to be an equivalent form where we had this function where we don't have the subtraction involving the square roots the square roots are fine as long as we don't have a subtraction that makes it a zero don't let this hide the fact that the thing that we really needed was the 0/0 part so here it was X divided by X we need those to cancel let's look at a very different technique for evaluating limits this is known as the squeeze theorem here we have three functions f of X G of X and H of X and we're going to let them satisfy this ending chain of inequalities and H of X is less than or equal to f of X which is less than or equal to G of X for some open interval containing a except possibly and a itself and then we need one additional thing we're going to suppose that the limit as X approaches a of f of X is equal to L and the limit as X approaches a of G of X is also equal to L so if we know that f of X is is trapped in between two things that are both going to L then the limit as X approaches a of f of X is also equal to L this is known as the squeeze theorem right because we're using H and G to squeeze F together towards a value so this can be really helpful on something like this limit the limit as X approaches 0 of x times the sine of 1 over X we looked at something very similar to the sine of 1 over X a couple of days ago this the sine of 1 over X is going to oscillate back and forth between negative 1 one and as X approaches zero but it'll oscillate with an increasing frequency and it will oscillate very very rapidly right around zero but it's being multiplied by something that's going to zero so we have something that is easily understood as X goes to zero and something that is very difficult to understand they may not actually sort of cancel each other out or they might let's see what happens this is also an easy example for using the squeeze theorem the squeeze theorem is very helpful for limits involving trig functions okay I can't plug in x equals zero because one over zero is undefined right so this is very different than the ones that we've looked at before so what we need to do is find bounds on this function X sine of 1 over X near zero as long as X is not zero the sine of 1 over X is in between negative 1 and 1 that's just what the sine function does it's never bigger than 1 it's never less than negative 1 but that's not actually the function that we want or if the function we want is x times the sine of 1 over X and but that's not too hard to figure out when X is positive we can just multiply by X so we get minus X is less than or equal to X sine of 1 over X less than or equal to X and when X is negative it reverses the order of the inequalities and but they're not going to reverse them again and so we get that X is less than or equal to X sine 1 over X which is less than or equal to minus X in either case we can combine these together and get that minus the absolute value of X is less than or equal to X sine 1 over X which is less than or equal to the absolute value of X look at the limits of those two bounding functions we can use the squeeze theorem now the limit as X approaches 0 of the absolute value of x is 0 and the limit as X approaches 0 of minus the absolute value of X is also 0 the pace of that function we were interested in X sine of 1 over X is bounded in between two things that both go to 0 so hey that must also go to 0 so the limit as X approaches 0 of X sine of 1 over X is equal to 0 so sometimes the answer is just going to be sort of a chain of equalities with some factoring involved like some of the first examples sometimes it's gonna be a little bit more of a paragraph here with some words explaining what we're doing so but either way and this is how we can evaluate some limits