this video is for those of you who might be going back to school and taking a calculus course let's start with limits so let's say we have some function f of x and it's equal to x squared minus four over x minus two what is the limit as x approaches two of that function so what exactly is a limit what does it do so basically what this question is asking is as x approaches 2 what does the y value or what does f of x approach now one of the first things you could try is direct substitution if we plug in 2 into the function it's going to be 2 squared minus 4 over 2 minus two two squared is four four minus four is zero and two minus two is zero zero divided by zero is undefined so that won't help us so what we can do instead is we can plug in a number that is very close to 2. let's try 1.9 so let's find the value of f of 1.9 so that's going to be 1.9 squared minus 4 over 1.9 minus 2. 1.9 squared minus 4 that's negative 0.39 1.9 minus 2 is negative 0.1 when you divide these two you're going to get positive 3.9 now what if we plug in a number that's even closer to 2 so let's say 1.99 what's going to happen well let's see what we're going to get so 1.99 squared minus 4. that's negative point zero three nine nine and one point nine this is supposed to be a two by the way not a four so let me just fix that 1.99 minus two that's equal to negative 0.01 and so let's divide these two numbers and this is equal to 3.99 so here's the question for you as x gets closer to two what happens to the y value or basically the value of the function notice that it gets closer and closer to four 3.99 is a lot closer to 4 than 3.9 is so therefore we can make the conclusion that the limit as x approaches 2 as we get closer and closer to 2 the value of this function approaches four and so this direct substitution technique can always work if you can't plug in the number directly you can plug in a number that's very close to this number and as you'll see the limit approaches a certain value now let's go back to the first problem it turns out that there's another way in which we could get the same answer so even though we can't plug in two directly because it's going to be undefined we can simplify the expression we can factor x squared minus 4. and it's a difference of perfect squares so it's going to be x plus two times x minus two so notice that we can cancel a factor we could cancel x minus two and so what we have now is the limit as x approaches two of x plus two at this point we can use direct substitution we're no longer going to have an undefined value so if we replace x with 2 it's going to be 2 plus 2 which is equal to 4 which confirms our previous answer so sometimes there are certain algebraic techniques that you can use to evaluate this limit other than using direct substitution with numbers that approaches two now that technique always works so it's a good fallback method but on a test you may have to show your work so you need to also learn the analytical methods with it as well so i'm going to give you four problems and i want you to evaluate the limit so let's start with this one what is the limit as x approaches 3 of x squared plus 5x minus 4. feel free to pause the video and work on this example now in this problem we could use direct substitution if we plug in 3 we're not going to get an undefined value we're not going to get a 0 in the denominator of a fraction so always check to see if direct substitution will work so this is going to be 3 squared plus 5 times 3 minus 4. 3 squared is 9. 5 times three is fifteen and fifteen minus four is eleven nine plus eleven is twenty and so this is the answer for this example now what about this one what is the limit as x approaches 3 of x squared minus 8x plus 15 divided by x minus 3. now if we try to use direct substitution if we plug in three three minus three is going to give us a zero in the bottom of the fraction which will make it undefined so therefore we don't want to do that let's see if we could factor the trinomial on top so what two numbers multiply to 15 but add to negative eight so we know that three times five is fifteen but we need it to add up to negative eight so we need to use negative three and negative five so therefore x squared minus eight x plus fifteen is equal to x minus three times x minus five and now we can cancel x minus three so we're left with the limit as x approaches three of x minus five by the way some teachers will require you that they want you to rewrite the limit expression until you replace x with three some teachers will take off points if you don't rewrite the limit expression so i just want to give you that warning so you don't lose points on a test so now let's plug in three so it's going to be three minus five at this point we don't need to rewrite the limit and this is equal to negative two and this is the answer now what about this one what is the limit as x approaches 4 given a complex fraction 1 over x minus 1 over 4 divided by x minus 4. now in this case we can't factor there's nothing to factor in this example but whenever you have a complex fraction here's my recommendation multiply the top and the bottom by the common denominator of the smaller fractions that you see so in this case let's multiply the top portion of the fraction by 4x and the bottom by 4x on top let's distribute 4x times 1 over x the x variables will cancel leaving behind four and don't forget to rewrite the limit expression so this is positive four and then if we multiply four x by one over four this time the fours will cancel leaving behind negative x on the bottom you don't need to change anything don't distribute the 4x just leave it the way it is notice that 4 minus x and x minus 4 are very similar they're not exactly the same at this point but there's something that we can do in the next step let's take out a negative one from four minus x let's factor it out if we factor out a negative one negative x changes to positive x and positive four will change to negative four so notice that now we can cancel x minus four so we're left with the limit as x approaches four of negative one divided by four x now we could substitute x with four and we don't need to rewrite the limit expression at this point so it's negative one divided by four times four and so the answer is negative one over sixteen so this is it now let's confirm that negative 1 over 16 is indeed the answer to this problem so if you're not sure you can always plug in a number that's close to 4. you can't plug in 4 directly because 4 minus 4 is 0. let's plug in a number that's close to 4. let's use 4.1 and then let's use 4.01 so let's find the value of f of point one so that's one over four point one minus one over four divided by four point one minus four now one over four point one minus one fourth that's equal to negative point zero zero six zero nine seven five and there's a six after that four point one minus four is just point one and so this gives you negative point zero six zero nine seven five six so let's keep this value for now now let's try f of 4.01 so that's 1 over 4.01 minus 1 4. over four point zero one minus four so the numerator has a value of negative point zero zero zero six two three four and the denominator is going to be point zero one so i'm gonna put this value up top since i'm out of space on the bottom so this is about negative point 0.0623 now let's say if we want to go further if we want to try f of 4.00 i'm not going to write it out i'm just going to type it in the calculator so this is going to be negative point zero six two four eight so notice that it's converging to a certain value here it was negative point zero six zero nine here it's a negative point zero six two three and now it's negative point zero six two four and i'm missing a zero it's supposed to be zero somewhere here so it's converging to a certain value which means that the limit exists now negative 1 over 16 if you type this in your calculator and convert it to a decimal this is negative point zero six two five as you can see these two answers are very close even though they're not exact but as x approaches the four as it gets closer and closer to four this value is converging to negative one over sixteen or negative point zero sixty five i keep forgetting the zero but that's supposed to be point zero six two five let's try one more problem evaluate the limit as x approaches nine given the function square root x minus three over x minus nine now if you're not sure what to do and if you have a multiple choice test you could find the value of f of 9.01 so that's close to 9. so let's see what this is equal to so this is 0.16662 so basically let's say 0.167 so the answer should be close to this value and if you want a better answer plug in 9.001 or even 9.001 the closer it is to 9 the more accurate the answer will be but if you have a free response test then you have to show your work whenever you have a square root inside a fraction what you want to do is you want to multiply the top and the bottom by the conjugate of the numerator in this case by the square root of x plus 3 as opposed to minus 3 so don't forget to change the sign and whatever you do to the top you must also do to the bottom now on the numerator we're going to foil the square root of x times the square root of x is simply x and then here we have positive 3 square root x and then let's multiply negative 3 by root x so that's negative 3 square root x and then negative 3 times 3 is negative 9. now what i recommend is not to foil these two keep it the way it is it's going to work out nicely if you don't foil x minus nine with the square root of x plus three and let's rewrite the limit expression so now we can cancel positive three square root x and negative three square x three plus negative three adds up to zero so what we have left over is x minus nine divided by x minus nine times the square root of x plus 3. so notice now that we could cancel x minus 9. so once you get rid of this x minus 9 factor you can now plug in 9 because this it prevents us from do so it will give us a 0 on the bottom so once that's gone now we can use direct substitution with the exact value so once we plug it in we don't need to rewrite the limit expression so all of this is equal to one over the square root of nine plus three and the square root of nine is three and three plus three is six so the final answer is one over six that's the exact answer now one over six as a decimal is basically point one six six six six repeating so you could say point one six which is basically point one six seven it's close to that so these two values they're in harmony with each other so whenever you're evaluating limits analytically see if you can either factor or if you have a complex fraction multiply the top and the bottom by the common denominator or if you have a fraction with a square root symbol inside of it multiply the top and the bottom by the conjugate of the square root factor now the next thing that you need to be able to do is you need to know how to evaluate a limit graphically so let's say if you want to find these four things what is the limit as x approaches negative 3 from the left side so that's a one-sided limit and what is the limit as x approaches negative 3 from the right side the right side is indicated by the positive superscript and what is the limit as x approaches negative three from either side and finally what's the value of f of negative three so these are common questions that you might see in a test when that'll went i lost my uh the cat caught my tongue when evaluating limits graphically so what is the value that f of x approaches as x approaches negative three from the left so first identify the value where x is negative three so that's this point here now if we follow the function from the left as we approach an x value of negative three we're going to be on this curve so we're approaching negative three and notice that the y value there is one so therefore as x approaches negative three from the left side f of x approaches one now what about from the right side so anywhere along this vertical line x is negative three so we're going to approach that x value starting from the right side and so it's going to take us to this vicinity where the y value is 2. so as x approaches negative three from the right side f of x is equal to two now what about the limit as x approaches negative three from either side if the left side and the right side if they differ then here's something important that you need to know the limit does not exist unless these two match the limit will not exist now what about the last one what is the value of f of negative 3 so when x is exactly negative 3 what is the function value so in this case we need to look for the closed circle at an x value of negative three and basically that closed circle has a value of negative one on the y axis so f of negative three is equal to negative one now whenever the limit doesn't exist you have a certain type of discontinuity this particular discontinuity is known as a jump discontinuity because there's a disconnect in a graph a jump discontinuity is also known as a non-removable discontinuity so let me give you some more questions relating to this topic so what is the limit as x approaches negative 1 from the left side and what is the limit as x approaches negative one from the right side now once you find those two values i want you to find the limit as x approaches negative one from either side and then find a value of f of negative one so anywhere along this vertical line x is equal to negative one so as we approach it as we approach an x value of negative one from the left side notice that the y value approaches negative three now as we approach it from the right side the y value is still negative three so notice that the left side and the right side have the same value therefore the limit exists and it's equal to negative three now what about f of negative one so along this line you want to look for the closed circle and here it is and it has a y value of negative two so f of negative one is negative two now this type of continuity or rather discontinuity where the limit exists but it doesn't equal to the function value is known as a whole and that type of discontinuity is known as a removable discontinuity so that's simply a whole now let's talk about another problem what is the limit as x approaches negative two from the left side and also what's the limit as x approaches negative two from the right side and also from either side and finally what's f of negative two so as x approaches negative 2 from the left side so here's where x is negative 2. as we approach from the left side notice that the graph goes higher and higher and higher so therefore it approaches positive infinity now as we approach negative two from the right side notice that it goes down towards negative infinity and these two they don't match so therefore the limit doesn't exist some teachers will also say that infinity because it's not a specific number they will define it as the limit not exist in two so just keep that in mind now what about f of negative two now there's no closed circle along that vertical line so therefore this value does not exist either you could say it's undefined because typically whenever you have let's say if you were to graph one over x minus two you're going to see a vertical asymptote at x minus two and at two if you plug it in you'd say it's undefined because f of two will equal one over zero so whenever the function doesn't exist due to a vertical line you could also say that it's undefined because that's why you have a vertical asymptote typically because there's a zero in the denominator so just keep that in mind now let's try one more set of problems what is the limit as x approaches 1 from the left one from the right one from either side and also let's find the value of f of one by the way you need to know what type of discontinuity this represents that's known as an infinite discontinuity and it's also a non-removable discontinuity now let's move on to the next problem as x approaches 1 from the left side notice that the y value approaches one and as x approaches one from the right side the y value is still negative one so because these two are the same as x approaches one from either side is going to be negative one the limit exists now f of one is also negative one because we have a closed circle at this point even if there's if there were no close circle if there was a straight line if we didn't have an open circle it would still be negative one now whenever all four of these values match up the function is continuous at x equals one whenever there's a mismatch it's discontinuous either it could be a whole it could be an infinite discontinuity or it could be a jump discontinuity when all four of these values match then we can say that the function is continuous at x equals one