so we continue section 2.3 with a few more limit techniques before moving on our next limit technique here will involve finding a common denominator so we all know very well that in order to add fractions we have to ensure that our fractions have the same denominator so as just a reminder in order to add two fractions or subtract two fractions we have to have common denominators and so we can do that here in this case we can multiply by the the first fraction by d over d and then the second fraction by b over b and now both of our denominators are b times d again we probably should say something about b and d of course not being equal to zero and so now these fractions add very simply as a d plus b c divided by b d and i have noticed a very small typo this is a d not a b but anyways this is how we add fractions we find a common denominator or subtract fractions and once we have our common denominator we can add or subtract the numerators so often the way that i do this is and certainly you do not have to follow this way but if you'll notice here our result our result is equivalent to multiplying across this way a times d keep our sign in between and then multiply across this way b times c and then divide by b times d on the bottom so often i see multiplication of frac or excuse me addition or subtraction of fractions as just a times d plus or minus b times c divided by b times d so this is often how i will approach the operation or the procedure the procedure of finding a common denominator so in this limit here i have the limit as x goes to 2 of 1 over x minus 1 over 2 divided by 2 minus x direct substitution of 2 of course is going to yield 0 over 0 so we cannot make we cannot yet make a conclusion on the basis of the indeterminate form about the existence of this limit or the value that this limit might turn out to be so notice in the top here i have two fractions so our approach here is going to be finding a common denominator so we want to combine those two so this is now going to read the limit as x goes to two so finding a common denominator we're going to have two times one minus x times one so we're going to have two minus x divide by the bottom multiplied x times 2 or 2x and that's all divided by 2 minus x so what we can do here effectively this is 2 on the bottom here this is 2 minus x over 1 if we want to think of it in those terms so to make the cancellation that we're going to get a little more obvious i'm going to write this as the limit as x goes to 2 this is going to be the same as 2 minus x divided by 2x right division of fractions is the same as multiplication by the reciprocal so this is equivalent to the previous step so 2 minus x divided by 2x times 1 over 2 minus x and now you should see we have a nice cancellation 2 minus x's that pair is going to be gone this is now just the limit as x approaches 2 of 1 over 2x and very simply now substitution of x equals 2 gives us our limit our limit comes to the value of 1 4. so through the process of finding a common denominator i was also to find a cancellation and once we have the cancellation our limit turns to be very simply just a matter of substituting x equals two and we're finished let's look here at an example where we might combine some of the techniques that we have now this is the limit as x approaches nine of x to the negative one-half minus one-third divided by x minus nine so let's just go ahead and rewrite some things the limit as x approaches nine so 1 excuse me x to the negative one-half is 1 over root x so i have 1 over root x divided by 1 3. that's all divided by x minus 9. so now in the numerator let's combine those two fractions by finding a common denominator so this is going to be 3 times 1 minus root x times 1 so 3 minus root x divide by 3x excuse me 3 root x right so that's 3 times 1 minus root x times 1 divided by 3 times root x that's all divided by x minus 9 which i'm going to go ahead and write as 1 over x minus 9. so we went ahead and switched our division of fractions to multiplication by the reciprocal so what might we do here so we still have an issue because plugging in nine nine minus nine is zero plugging in nine three minus root nine is zero so this all comes out to still being zero over zero so how about we multiply by a conjugate i see a square root here very clearly the conjugate will be three plus root x so let's just multiply by the conjugate of that numerator expression which is just going to be 3 plus root x and so now what we have is going to be the limit as x approaches 9. so on the top multiplying our conjugates 3 times 3 is going to be 9. root x times root x is going to be x so root x times root x is going to be x so i have 9 minus x on the top and then i have that still being multiplied times 1. on the bottom here we'll just leave this unfactored or excuse me unexpanded so leave it as is so i have 3 root x times x minus 9 and then times our conjugate term 3 plus root x so notice we almost have a cancellation here but not quite 9 minus x is close to x minus 9 but these are off by a factor of negative 1 or off by a multiple of negative 1. so what i'd like to do here is rewrite 9 minus x this way so 9 minus x is the same as negative x minus 9. so hopefully you will agree with that distribute the negative one i have negative x distribute the negative one i have positive nine so let's go ahead and make that change i'll do this without rewriting our limit completely that nine minus x in the top is now negative x minus 9. so that's an equivalent way right these these are equal i'm just replacing one with the other and now we get what we want we cancel but we're not quite canceling to a positive one we now have the introduction of this negative one that's going to stay along for the ride so in general you might notice this you might recognize this pattern here x minus y divided by y minus x will always cancel and equal negative one right so long as y and x are not the same so let's rewrite this one more time we now have this expression we have the limit as x approaches nine all that's left over on the top is a negative one and on the bottom i have this expression three root x times the conjugate three plus root x all right so going back all i have now is negative 1 divided by 3 root x times 3 plus root x and we are in good shape to complete this limit by direct substitution so the negative 1 is going to come along there's nothing to plug in in the top on the bottom we'll plug in x equals 9 so 3 root 9 is going to be 3 times 3 that's going to be 9. here i'm going to have 3 plus root 9 that's going to be 3 plus 3 that's going to be 6. so this is going to be 9 times 6. this limit evaluates to negative 1 divided by 54. we want to be able to evaluate limits involving piecewise functions so a piecewise function is as the name implies it's a function defined in one or more or more likely two or more pieces here this piecewise function is called f and here's how it's defined the function out the output of the function so f of x in other words the y value is equal to zero when x is less than or equal to negative five we're equal to the square root of 25 minus x squared so long as x is between negative 5 and positive 5 and then we're equal to 3x if x is larger than or equal to 5. so if i'd like to find the limit as x approaches five so again for the limit as x approaches five well notice here at x equals five we have a transition so when we are less than five we're equal to this piece root 25 minus x squared and we're equal to 5 we're going to be equal to 3x so again for the two-sided limit i need to check a couple of things i need the limit as x goes to 5 from the right all right and as x goes to 5 from the right so the right side of 5 means we're larger than 5 so we're going to be on this piece of this function 3x so of course this limit is going to equal 15. so i've run out of room on the edge there so the limit as x goes to 5 on the left is equal to 15 as we go to five excuse me at five on the right as we go to five on the left well five from the left is mean it is telling us we're approaching from values smaller than five and being smaller than 5 we're going to be on this piece the root of 25 minus x squared all right so plugging in five very clearly you're going to see a value of zero so we'll have 25 minus 25 under the root root zero is zero now since the left hand limit is zero the right hand limit is 15 our two-sided limit the limit as x approaches five we will say does not exist right of course since 15 does not equal zero the next limit we have is the limit as x goes to negative five so as we approach negative five on the left hand side we're equal to zero and as we approach negative five on the right hand side or equal to this piece root 25 minus x squared so for the limit as x goes to negative 5 i need to look at two limits so let's look at the limit as x goes to negative 5 on the right so negative five on the right means that we are approaching from values larger than negative five so values larger than negative five we'll put us here and we'll be on this piece root 25 minus x squared so plugging in negative 5 well of course negative 5 squared is 25 we'll again see that this limit is zero now the limit as x goes to negative five on the left well we're approaching from values smaller than negative five smaller than negative 5 puts us here on this constant y value of 0 so this is just the limit as x goes to negative 5 from the left of 0 well there's no variable it's constant so it's also zero so here we see since the left-hand limit is zero the right-hand limit is zero obviously zero equals zero we can now say our limit as x approaches negative five of our piecewise function is going to exist and it's going to be equal to zero so given a piecewise function you want to be able to determine some limits let's look at this example involving a piecewise function the question here says determine a value for k so k being some real number for which the limit as x goes to 2 exists and state the value of the limit so our piecewise function is the line 3x plus k when x is less than or equal to 2 and x minus 2 when x is larger than 2. so what has to happen for the limit to exist for the limit to exist we have to have the right hand limit equal to the left hand limit all right so as we approach negative 2 from the right hand side so again from values larger i'm sorry these should be positive 2. all right so as we approach 2 from the right we're approaching from values larger than 2 so that's going to put us on this piece of our function x minus 2. and so logically on the left-hand approach to 2 we're approaching from values smaller than 2 that puts us on this segment 3x plus k so evaluating these limits is simply now a matter of plugging in 2 so of course 2 plugged in on the right hand side limit is going to be 0 2 plugged into the left hand side limit is going to be 3 times 2 so 6 plus k and solving for k k will be required to be equal to negative 6. so in that case what will the limit as x approaches 2 equal the limit as x goes to 2 for this function well left and right have to be equal so right 0 we've solved for k to make that happen on the left hand side anyways the limit as x goes to 2 will have to be 0 in this case so the last technique from section 2.3 involves what we call the squeeze theorem and the idea of the squeeze theorem is very simple so what is required for the squeeze theorem is that we have three different functions f g and h and we have this ordering of the functions around some number c we want f of x to be less than or equal to g of x and g of x to be less than or equal to h of x so we want this inequality to hold for all x near the value of c except possibly at c itself so this inequality could break down at c but again that's okay we don't care about what happens at c itself because we are talking about a limit so this inequality holds near the value c furthermore suppose we also have the next two further conditions the limit as x goes to c of f of x is equal to the limit as x goes to c of h of x and both of those are equal to some number l so if the limit as x goes to c of f of x is l on one side the limit as x goes to c of h of x is equal to l on the other side then the only possibility that we can have this limit as x goes to c of g of x in the middle is also going to be required to be equal to that same value of l so very simply suppose i asked you to find all numbers x for which this inequality holds well the only number that i can place here in the middle and make this inequality remain true is going to be 2 itself so the idea of the squeeze theorem is the same if the limit on the left is the same as the limit on the right then the requirement is going to be that the limit in the middle is also the same so that's the idea of the squeeze theorem in words here's a graphical representation of the squeeze theorem in black here i have the function f of x equals x squared times sine of 1 over x now as x goes to 0 i cannot evaluate this because 1 over 0 plugged in here is going to be undefined and so in calculus i cannot say that zero squared times something undefined is going to be zero but what you'll notice here is that in blue this is the curve y equals x squared in red this is the curve y equals negative x squared and so what we're seeing is that this function is squeezed in between positive x squared and the negative x squared and so what we're going to be able to conclude as x goes to 0 we have this function x squared on top its limit is going to be 0 we have this function on the bottom negative x squared its limit is also going to be zero so i'll be able to conclude that this limit as x goes to zero of this function is also going to be equal to zero so again the name sort of implies what's happening our function of interest is being squeezed in between two others and the hope is that the others are going to have easily calculated limits this theorem is sometimes called the sandwich theorem in other textbooks i've even seen in older textbooks it's called the pinching theorem our textbook refers to it as the squeeze theorem so that's what we will stick with so in this example suppose i'm given that 2 minus x squared is less than or equal to g of x and g of x is less than or equal to 2 cosine of x and this is happening for all values of x so i'd like to find the limit as x goes to 0 of this function g of x so i don't really know much about this function g of x but i do know that it's squeezed between these two functions 2 minus x squared and 2 cosine of x in general you may not be told that you need to use the squeeze theorem but this presence of an inequality should immediately signal squeeze theorem so finding this limit what i'm going to do is i'm going to apply the limit across the inequality so we have the limit as x goes to 0 of 2 minus x squared that's going to be less than or equal to the limit as x goes to 0 of this unknown function g and that's going to be less than or equal to the limit as x goes to 0 of this function 2 cosine x and so the limit on the left and the right are easily evaluated and they're evaluated immediately by direct substitution so plugging in 0 on this left side limit of our inequality of course 2 minus 0 squared is going to be 2 so what i have is 2 less than or equal to the limit as x goes to zero of our unknown function g and on the right hand side of our inequality plugging in zero the cosine of zero is one two times one is of course two so the left hand or i should say more clearly the left side of the inequality the limit is 2 the right side of the inequality the limit is 2. so if our limit as x goes to 0 must satisfy this inequality there's only one potential choice and the choice is that this limit in the center is also equal to two so we'll justify this by saying by the squeeze theorem what we have is that the limit as x goes to zero of g of x is equal to two so this is a very simple instance of applying the squeeze theorem so in other cases we may need to build the inequality so we may not have the inequality given to us we might need to build it ourselves this typically happens when we have trig functions in particular sine and cosine so you want to remember the range of the sine and the cosine functions so you should know that the range of the sine and the cosine functions is this interval negative 1 to positive 1. so in other words all values of sine are larger than or equal to negative 1 but less than or equal to positive 1 and the same goes for cosine and this is going to be true regardless of what's shoved into these two functions so more generally if we have an arbitrary function call it f the sine of f of x is still going to satisfy this inequality so we're still going to be larger than or equal to negative 1 but less than or equal to positive 1. so regardless of what's shoved into the sine function this is going to always hold true so these are going to often be our our starting points for developing a squeeze theorem argument so let's take a look here i'd like to evaluate the limit as x goes to 0 on the right of x times the cosine of 1 over x so again the issue here with direct substitution is 1 over 0 is undefined so i there's nothing i can say about the cosine of one over zero so i'd like to develop an inequality here so as we just mentioned the cosine of anything satisfies this inequality so in this case we have the cosine of one over x is larger than or equal to negative one yet less than or equal to positive one so the goal here is to sort of is to build this expression this is what i want to show up in the middle because this is the expression that's giving me difficulty so here we're just looking at x approaching 0 from the right so we are looking at x approaching zero from values larger than zero so what we're going to do is we're going to multiply our inequality by x and since x is positive we do not have to worry about the direction of our inequality being disturbed so as if we multiply across the inequality by x what we're going to have is this expression we're going to have negative x is less than or equal to x times the cosine of 1 over x and that's going to be less than or equal to 1 times x which is of course just going to be positive x so now i have this expression in the center which is exactly the expression that we're wanting to find the limit of as x goes to zero on the right so we'll apply the squeeze theorem i'm going to apply the limit as x goes to zero on the right across the inequality so again the thing in the middle is the one that we can't evaluate directly and on the right hand side limit as x goes to zero on the right of just x well on the left and the right of our inequality these are simply direct substitutions of zero negative zero is of course zero so i have this inequality developing 0 is less than or equal to the limit as x goes to 0 on the right x times the cosine of 1 over x and on the right hand side again this is just a linear function effectively we can plug in 0 into a linear function always and so this is going to come out to being 0 on the right so what do we have our unknown limit is greater than or equal to zero but also less than or equal to zero so the only conclusion is we'll say by the squeeze theorem and here i will abbreviate i'll abbreviate st the limit as x goes to zero on the right of x times the cosine of one over x is going to be equal to zero we could develop a similar similar argument for x going to zero on the left however in that case because x going to zero in the on the left represents values less than zero multiplying our inequality by a negative number or the representation of a negative number is going to turn around the signs of our inequality the same process will apply we should still find that the limit as x goes to zero on the left is also zero but we do have to take a little bit more care because multiplying an inequality by a negative number or the representation of a negative number requires the direction of our inequality to change so let's work one last example in this section involving the squeeze theorem to work this last example we need to add a fact to our knowledge base about exponential functions so first definition a function f is said to be strictly increasing if a less than b means that f of a is less than f b for all a and b in the domain of this function f so in other words if i have two input values and one input value is less than the other then if a function is strictly increasing that means that the output values corresponding to these inputs are going to be less than each other so this pertains to exponential functions exponential functions with base b larger than 0 are strictly increasing so you can see that when you see the graph of an exponential when our base is larger than 0. so what this means is that we can apply a inverse excuse me we can apply an exponential function across an inequality and that inequality is going to hold and that's due to this this due to this behavior of the exponential function with base b larger than 0 being strictly increasing so for instance e to the x is a strictly increasing function so what we can say is that if we know that one-half is less than three-fourths which is of course true then we are automatically guaranteed that e to the one-half must be less than e to the three-fourths so for exponential functions with a base larger than zero i know that i can apply such a function across an inequality and it will not disturb the inequality that i have written so we'll see that here on this example i'd like to prove that the limit as x goes to 0 on the right of the root of x times e to the sine of pi over x is equal to zero again direct substitution fails because pi divided by zero is undefined so let's again build our inequality let's start from that fact involving the sine function of course i can say that negative 1 is less than or equal to the sine of pi over x and that's going to be less than or equal to positive 1. all right so there's the sine of pi over x now notice here i have this exponential involved so i can begin to reach this expression that i have by exponentiating each part of our inequality and again this is exponential base e e is larger than zero so i can apply i can exponentiate across this inequality and we're not going to lose the signs that we have here so this is going to be the same thing as e to the minus 1 being less than or equal to e to the sine of pi over x and that's less than or equal to e to the first power which is just which is of course just e all right we're almost to this expression i now have a root x that i'd like to introduce square roots are always positive especially since we are away from zero as we approach zero the square root of a number is always positive so i can now multiply everything by the square root of x and we still have the inequality and on the right hand side i have root x times e will just stay consistent e root x so this is exactly what i have in the middle that i that i was given so root x times e to the sine of pi over x so we have this inequality now notice that the left and right hand or excuse me the left side of our inequality and the right side of our inequality are going to be very simple to evaluate so we're going to apply the limit across the inequality so i have e to the negative first power times root x less than or equal to the limit as x goes to 0 of this expression root x times e to the sine of pi over x and then that's going to be less than or equal to the limit as x goes to 0 of e root x so now we're ready to apply the squeeze theorem by substituting on the left and right side of our inequality so here plugging in zero root zero 0 e to the negative first power is just a number so anything of course times 0 is still 0. in the middle we have the limit of interest root x times e to the sine of pi over x and on the right hand side plugging in 0 e times 0 is still 0. so here we have the conclusion again by the squeeze theorem the limit as x goes to zero on the right we have now proved root x times e to the sine of pi over x is equal to zero