in section 2.3 we're going to discuss techniques for computing limits and in my opinion this is the most important section in chapter two because the techniques in this section we will also be using in chapter three when we are developing the derivative by definition so the techniques in this section are going to show up again in chapter three so we'll actually take two classes for section 2.3 in order to spend some considerable amount of time thinking about the various limit techniques that we have available so our goal is to be able to work a limit by hand and we've seen previously we can work limits graphically we can also work limits numerically but those have some limitations and so going forward we we want to be able to determine a limit uh exactly um and so we're going to be developing in this section some some algebraic techniques for computing limits so first off we introduce some limit laws so i'm going to assume that i have these two particular limits the limit as x approaches c of f x in the limit as x approaches c of g of x we're going to assume those limits exist and we're also going to make an assumption here or a declaration that's this number k that we'll discuss down below is a real number so here are our limit laws so a limit can be split across a sum or a difference so if i have a limit if i have a limit given to me and the inside of this inside of this limit has pieces that are being added and subtracted what we can do is we can split that into individual limits addressing each term so if i have the limit as x approaches c of f of x plus or minus g of x i can equivalently write the limit as x approaches c of f of x plus or minus the limit as x approaches c of g of x i also have a property of being able to remove constant multiples outside of the limit so here again k is just some real number if i have the limit as x approaches c of k times f of x that constant k can be pulled out to the front so these two properties together are called linearity so in mathematics when we talk about we talk about linear operators these two properties are what define a linear operator so i can split the operation across a sum or difference and i can pull constant multiples out of the operation so these two right here are what we call the the linearity properties if i have the limit of a product so the limit as x approaches c of f x times g of x that's equivalent to the product of the limits so i can take the limit as x approaches c of f of x multiply by the limit as x approaches c of g of x and if i have a quotient involved in my limit i can apply the limit to each piece of the quotient so the limit of the numerator divided by the limit of the denominator and of course since we're talking about division we'll require that lower limit to not be equal to zero so these this collection of laws here are what we call the limit laws so let's apply these here let's suppose that the limit as x approaches one of f of x is eight the limit as x approaches one of g of x is three and the limit as x approaches one of h of x is two so i'd like to use my limit laws to evaluate this limit the limit as x approaches one of f of x times g of x divided by g of x minus three times h of x so we'll apply our limit laws so we have the limit here of a quotient so this is equivalent to the quotient of the limits so we'll apply the limit to the top and the bottom individually so i have the limit as x approaches 1 of this product on the top and then this difference g of x minus 3 times h of x and so now we know on the top the limit of a product is the product of the limits so i have the limit as x approaches 1 f of x times the limit as x approaches one of g of x and on the bottom we can split this limit into two pieces due to this subtraction so we'll have the limit as x approaches one of g of x and on the second piece i'll go ahead and pull the constant three out as well so we'll have minus three times the limit as x approaches one of h of x so we're ready to finish this now what where what were we given well we had the limit as x approaches 1 of f of x that was given to us as being equal to 8. as x approaches 1 for g of x we have the limit being three so on the top we're going to have eight times three again here on the bottom as x approaches one for g of x the limit is given as three and on the bottom this limit involving h as x approaches one is going to be equal to two so on the top eight times three is of course twenty four on the bottom what do i have i have three and here i'm gonna have three times two which is six so three minus six is of course negative three and so our limit is going to simplify to negative eight so the first technique that we'll discuss is the technique of direct substitution so the idea of direct substitution is very simply if i'm able to plug in a number into the function and in particular the number that we're approaching with regard to our limit if i'm able to plug that number in and the result is a real number then we're finished and there's nothing else to do so this method of direct substitution or this technique of direct substitution is often the first thing that we will try when we are beginning the process of evaluating a limit so if direct substitution gives me a real number then our work is finished and we've evaluated the limit so direct substitution will work when the point we are approaching with regard to our limit is contained within the domain of the function so this ties into something uh that we will discuss later on in chapter two on that topic being the topic of continuity and continuous functions so for now i'm just going to make the observation that these functions polynomials exponentials the absolute value function the sine and the cosine functions all have a domain of all real numbers and so functions that are involving just some combination of these will always be able to be evaluated through this technique of direct substitution there are other functions that have domain all real numbers but for the sake of this particular this particular lesson we're just going to focus on on these five right here so let's take a look at this technique of direct substitution here i have the limit as x goes to 0 of 2x minus 1 divided by x minus 2. so again our first our first idea for evaluating a limit is going to be direct substitution and if direct substitution gives me a real number we are going to be finished so we're just going to take that point of zero plug it into this expression and if the result is a real number then we are finished so plugging in zero i'm gonna have two times zero minus one divided by zero minus two so on the top very simply that's going to be zero minus one which is of course negative one on the bottom zero minus two which is of course negative two so negative one divided by negative two is of course positive one half positive one half is certainly a real number and so we have finished evaluating this limit this limit is equal to one half the next limit i have here is involving a polynomial in fact this is a quadratic polynomial or a degree two polynomial so let's just plug in one so plugging in one i have two times one squared plus three times one plus one so of course one squared is one three times one is three so two plus three plus 1 is of course 6. again 6 is a real number so we're finished our limit is equal to 6. and then lastly here i have the limit as x approaches 2 of e to the 1 half x minus 1 plus the cosine of x pi so exponential function cosine function domain is all real numbers i certainly expect direct substitution to work so we're going to plug in 2 so we're going to have e to the half times two minus one plus the cosine of two pi so this exponential expression a half of two is one so i have e to the one minus one the cosine of two pi you should know from the unit circle is one which is also the cosine of zero now one minus one is of course zero so i have e to the zero plus one and any exponent of zero results in one so we have a one plus one so our limit evaluates out to two so this is the technique of direct substitution if i get a real number everything is all well and good and we're done however as you might imagine there's not really a whole lot to talk about if direct substitution is always going to work so again not much of a surprise direct substitution is not always going to work so in particular direct substitution may give us that 0 over 0 indeterminate form that we were talking about in section 2.1 so the zero over zero indeterminate form we have to be very careful zero over zero we talked about is not equal to one and zero over 0 does not mean that the limit does not exist when we apply direct substitution and get that 0 over 0 that's a message and that message says you've got more work to do okay so we have to be very careful zero over zero will never be the result of a limit and we can never make a conclusion about the existence of a limit or the actual value that a limit will turn out to be simply based on that zero over zero expression and that's why we call it an indeterminate form and so let's look at some of the cases where we will come up with that zero over zero and some techniques where we can work around it in order to evaluate our limit so the first technique that we'll use is a very simple technique of just factoring and canceling terms so back in section 2.1 we claimed that this limit was going to be equal to 6 we didn't necessarily justify it but now we're able to justify it we can now show that this limit is going to be equal to 6. so again how will we do that we make this approach of factoring and canceling terms so you'll notice that the numerator is a difference of squares and a difference of squares factors this way so we have x plus three times x minus three and on the bottom we still have this x minus three and so here very nicely we're noticing that x minus three over x minus three cancels so what we're left over with is the limit as x approaches three of x plus three now notice we are in great position to simply apply a substitution and be finished so plugging in 3 of course 3 plus 3 is going to give me 6. so in general the idea for these types of limits in this section the goal is to get down to a position where direct substitution no longer presents an issue and so this is one of our main techniques this fact this this technique of factoring and canceling and so here we claimed the other day this limit was six and here's how we show it now here we have the limit as x approaches two of x squared minus seven x plus ten divided by x minus two so plugging in two two squared is going to be four four minus seven times 2 is going to be 4 minus 14 which is negative 10 negative 10 plus 10 is of course 0 on the bottom 2 minus 2 is of course 0. so we'll sometimes write this out in quotes because 0 over 0 is not a number we want to be careful to not write equals so notice we have our 0 over 0 indeterminate form so there's my message my message says you have more work to do in order to finish this limit so we'll apply that technique of factoring and canceling so the clue here is you know i have a quadratic quadratics tend to be able not all quadratics but most quadratics that we will come across in these settings tend to be able to be factored and part of your factor the clues there's a clue here in the bottom the goal here is to make this cancel so this top polynomial this top quadratic factors into x minus two and x minus five and then we divide by x minus two so you can foil that top out to check x squared minus five x minus two x there's minus seven x negative 2 times 5 negative 5 is positive 10. so we have our cancellation of terms so the limit as x approaches 2 of x minus 5 is what we're left over with plugging in 2 2 minus 5 is of course negative 3. so by our technique of factoring and canceling our limit for this particular function that we were given is negative three so let's look at another involving a factoring and cancellation of terms so here i have the limit as x approaches one of two x squared minus three x plus one divided by x squared plus two x minus three so if you plug in one on the top and on the bottom you should notice that we have zero over zero so we have the indeterminate form so we wanna continue working this down to an expression that is more palatable the one that's more easier to work with one that allows us in effect to just complete this by direct substitution so the bottom is going to factor like this so the bottom factors into x plus three and x minus one so we can check that x squared minus three x excuse me x squared minus an x plus three x is the plus two and then three times negative one is the minus three and the top will factor the top will factor into 2x minus 1 and x minus 1. so this is exactly what we wanted to happen we have a cancellation now so our limit becomes the limit as x approaches one of two x minus one divided by x plus three so on the top we will plug in one two times one is two two minus one is one on the bottom we'll plug in one one plus three is of course four and so our limit is one fourth so your skill from algebra of factoring polynomials in particular quadratics mostly we don't really see the need often to factor cubics but sometimes they do pop up but your skills from algebra and factoring quadratics and polynomials will often be important in calculus so in this particular instance we need those techniques of being able to factor expressions polynomial expressions in order to be able to evaluate this limit right another limit here that's going to involve some factoring and cancellation is the limit as x goes to zero of three x squared plus five x divided by x squared plus two x very quickly you should see substitution by of zero gives you zero over zero so on the top you should notice that we can factor out an x this will be x times three x plus five same can happen on the bottom i have x times x plus 2. so this is what we want to see we want something to cancel so now we're just left with the limit as x approaches 0 of 3x plus five divide by x plus two so we are in good shape to make our final step of direct substitution plugging in zero this is of course going to be zero plus five and zero plus two on the bottom so our limit is 5 over 2. a third technique is the technique of multiplying by the conjugate so in mathematics expressions of this form a plus b and a minus b are called conjugates so these terms a and b remain the same and i change the sign between them and we call expressions like that conjugates so the conjugate of a plus b is a minus b the conjugate of a minus b is a plus b now conjugates have a nice property when we multiply them the cross terms are always going to cancel so when i take a minus b multiplied by a plus b we have to foil but we will always see these cross terms canceling so when i multiply conjugates i can actually ignore the foiling technique and what we'll always end up with is the conjugate a minus the conjugates a minus b times a plus b always becomes a squared minus b squared so this technique of multiplying by the conjugate will mostly be used when you see a square root involved in your limit and you also have an indeterminate form of zero over zero so the the benefit here is that when we multiply by the conjugate we're typically going to be multiplying a square root times the square root and when we do that what happens is of course this is x to the one half times x to the one-half the powers are going to add a half plus a half is of course one and x to the first power is of course x so multiplying the square root of x by the square root of x has the effect of eliminating the root and in some limits where we have square roots involved that elimination of the root may be very advantageous in our process of evaluating the limit so in this example let's just multiply this expression by its conjugate so the root of x minus 1 plus the root of x we're going to multiply by its conjugate so we have the root of x plus 1 plus the root of x times the conjugate so the root of x minus 1 minus the root of x so again when we multiply conjugates again we should foil when we have expressions when we have binomials multiplied by each other but i don't need to foil here because we have a conjugate being multiplied we have conjugates being multiplied together so what we're going to have is we're going to have the front two terms so the root of x minus one times the root of x minus 1 and then minus the root of x times the root of x so we've multiplied this is our a times a so that's a squared at the front and then minus b squared which is root x times root x so here this is going to drop the root we're going to have x minus 1 minus x so we're going to have x minus 1 minus x simplifying to negative one so this expression multiplied by its conjugate simplifies down to negative one let's take a look at this limit we have the limit as h approaches zero our limit expression our function here is the root of five h plus four take away two divide by h so plugging in zero you should notice that we have the root of four which is two so two minus two on the top divided by zero so we have this indeterminate form zero over zero so often when we see this indeterminate form involved with a square root the technique is going to be to multiply by the conjugate in this case the conjugate of the numerator so the conjugate of the numerator is the root of five h plus four plus two and we have to do that on top and bottom because we want to make sure that we're just multiplying by one and we're not changing the essence of this expression so on the top we can simplify the limit as h goes to zero again conjugates we don't need to actually foil we can just multiply the first two terms so we're going to have 5h plus 4 times 5h plus 4 under both of the square root and then minus 2 times 2 so we have minus 4. so these two terms multiplied subtract these two terms multiplied now on the bottom we will typically not make a cancellation excuse me we will not make an expansion so let's just keep this as it is so we have h times the conjugate so what happens here multiplying the two terms involving the root on top makes the root go away so we have 5h plus 4 minus 4 divided by h times the conjugate so what's going to happen 4 minus 4 obviously is going to be zero so let's write another step here we're gonna have the limit as h goes to zero i've got five h left over on top divided by h times the conjugate root five h plus four plus two now notice we have a further cancellation these h's are going to drop and we're ready to finish so on the top i'm just left over with a five on the bottom we're going to plug in zero so we'll have root zero plus four which is root four root four is two two plus two is of course four this limit evaluates to five divided by four so here this technique of multiplying by the conjugate gave us a couple of cancellations and it brought us to a point where we were able to finish by the process or the technique of direct substitution so let's look at one more example for today here we have the limit as x goes to four and i have z minus excuse me z goes to four and i have z minus four divided by root z minus two you should notice that plugging in four gives me zero over zero now i also have this square root involved so a first thing to try or a first instinct for these types of limits is to multiply by the conjugate the conjugate is root z plus two so multiply by the conjugate on top and bottom so i have the limit as z approaches four on the top i have z minus four times the root of z plus two now on the bottom we're going to have the multiplication of the conjugates so that means that we can do a squared minus b squared so root z times root z minus b squared two times two so perhaps you're seeing what's going to happen here i've got the limit now as z approaches four on the top i have z minus four times root z plus two now on the bottom root z times root z is z so i have z minus four and bingo there we go we have the cancellation of the z minus fours now we're ready to just substitute in four here plugging in four i have the root of four plus two the root of four is two two plus two is of course four and so our limit evaluates to four so in general when you see the square root the along with the indeterminate form the instinct will be multiplying by the conjugate typically we will expand when we multiply the conjugates but the terms that are not conjugates typically we do not expand because we're going to see a cancellation somewhere and so if you make an expansion here it's going to be difficult to see the cancellation so typically we multiply the conjugates to simplify we do not multiply the terms that are non-conjugate so this is where we will end the first part of section 2.3 in the second part we'll look at a couple more techniques for evaluating limits before moving on into section 2.4