in this video we're just going to go over a basic introduction into limits and how to evaluate them analytically and graphically so here's a simple example let's say if we want to find the limit as x approaches two of the function x squared minus four divided by x minus two so how can we do so well one way is to use direct substitution if we plug in two notice what will happen two squared is four four minus four is zero so zero over zero is undefined which we don't know what value that represents now sometimes you could find the limit by plugging a value that's close to two and that's what you want to do you want to plug in a number that's close to two but not exactly two so for example let's call this f of x so let's calculate f of 1.9 and let's see what's going to happen actually let's make it 2.1 so let's get a positive answer instead of a negative one two point one squared minus 4 that's about 0.41 and 2.1 minus 2 is 0.1 so this is going to be 4.1 now what if we pick a value that's even closer to 2 for example let's try 2.01 so if you type this in the way you see it in your calculator you may have to put this in parenthesis you should get 4.01 so notice what's happening as we get closer and closer to two the limit approaches four so we could therefore say that the limit as x approaches two of this function is equal to four and this technique works for any limit as long as you plug in a number that's very close to whatever this number is but not exactly that number if the limit exists it's going to converge to a certain value now sometimes you have to use other techniques to get the answer in this particular example we could factor x squared minus four you can write it as x plus two times x minus two now we need to rewrite the limit expression until we replace x with two now notice that we can cancel x minus two so when this term is gone we can now use direct substitution because the x minus two factor was giving us a zero in the denominator which we don't want so now all we need to do is find the limit as x approaches two of x plus two so now we can replace x with two and two plus two is four and so that's the limit it approaches a value of 4. now let's look at another example what is the limit as x approaches 5 of x squared plus two x minus four so notice that we don't have a fraction we're not going to get a zero in the denominator so for a question like this you can use direct substitution so all you got to do is plug in five so it's going to be five squared plus two times five minus four so that's 25 plus 10 minus four and 25 plus 10 is 35. so the limit is going to be 31. and so that's it for that example but now what about this one what is the limit as x approaches 3 of x cubed minus 27 over x minus 3. now if we try to plug in 3 it's going to be 0 over 0 so we don't want to do that in this case if you have a fraction like this see if you can factor the expression so how can we factor x cubed minus 27 so what we have is a difference of cubes and whenever you see that you can use this formula aq minus b cubed is a minus b times a squared plus a b plus b squared so in our example a to the third is like x to the third and b to the third is 27. so a is the cube root of x cubed which is x and b is the cube root of 27 which is three so this is going to be x minus 3 and then a squared that's x squared and then plus a b so that's 3 times x and then plus b squared or 3 squared which is 9. so now we can cancel the factor x minus three so what we have left over is the limit as x approaches three of x squared plus three x plus nine so at this point we now can use direct substitution so it's three squared plus three times three plus nine which is nine plus nine plus nine adding nine three times is basically multiplying nine by three and so this limit is equal to twenty-seven now here's another problem that you can work on so what is the limit as x approaches three of one over x minus one over three divided by x minus three so for these examples feel free to pause the video if you want to and try these problems so in this example we have a complex fraction so what do you do in a situation like this if you get a complex fraction what i recommend is to multiply the top and the bottom by the common denominator of those two fractions that is by x and by 3. so i'm going to multiply the top and the bottom by 3x so if we multiply 3x by 1 over x the x variables will cancel and so what we're going to have left over is simply 3 and if we multiply 3x by 1 over 3 the 3s will cancel leaving behind x but there's a negative sign in front of it now for the terms on the bottom i'm going to leave it in its factored form so notice that 3 minus x and x minus 3 are very similar if you see a situation like this factor out a negative one if we take out a negative one the negative x will change to positive x and positive three will change to negative three so notice that we can cancel the x minus 3 factor at this point and so what we have left over is the limit as x approaches 3 of negative 1 over three x so now we can use direct substitution so let's replace x with three so it's going to be three times three which is nine so the final answer is negative one divided by nine so now you know how to evaluate limits that are associated with complex fractions so here's another example what is the limit as x approaches nine of square root x minus three over x minus nine so what should we do if we're dealing with square roots what i recommend is to multiply the top and the bottom by the conjugate of the expression that has the square root so the conjugate of square root x minus 3 is square root x plus 3. so in the numerator we need to foil so the square root of x times the square root of x is the square root of x squared which is simply x and then we have the square root of x times three so that's going to be plus three square root x and then these two will form negative three square root x and finally we have negative three times positive three which is negative nine now on the bottom i'm not going to foil the two expressions i'm gonna leave it the way it is because my goal is to get rid of the x minus nine i want to cancel it so now negative three and positive three add up to zero so what we have left over is the limit as x approaches 9 of x minus 9 divided by x minus 9 times the square root of x plus 3. so now at this point notice that we can cancel x minus nine and so what we have left over is the limit as x approaches nine of one over square root x plus three so now what we can do is replace x with nine so i'm just going to continue up here so it's going to be 1 over square root 9 plus 3 and the square root of 9 is 3 and 3 plus three is six so the final answer is one over six now let's look at this example what is the limit as x approaches four of the expression one over square root x minus one over two divided by x minus four so this time we have a complex fraction with radicals that means we need to multiply the top and the bottom not only by the common denominator but also by the conjugate but let's start with a common denominator so i'm going to multiply the top and the bottom by these two that is by 2 square root x so when i multiply 1 over square root x times 2 square root x the square root x terms will cancel leaving behind positive two so i have the limit as x approaches four with a two on top and if i multiply these two the twos will cancel leaving behind the square root of x and so on the bottom i have 2 square root x times x minus 4. so now my next step is to multiply the top and the bottom by the conjugate of the radical expression that is by 2 plus square root x now i'm only going to foil the top part not the bottom so we have 2 times 2 which is 4 and then we have 2 times the square root of x and then these two that's going to be negative 2 square root x and negative square root x times positive square root x is going to be negative square root x squared which is negative x and in the denominator don't foil just rewrite what you have if you follow these steps it won't be that difficult so now 2 and negative 2 adds up to 0. so what we have left over is the limit as x approaches 4 of 4 minus x divided by all the stuff that we have on the bottom now what we're going to do at this point is we're going to factor out a negative one so this is now the limit as x approaches four so negative one and this is going to change to positive x and plus 4 is going to change to negative 4 just like we did before so now notice that we can cancel the x plus four terms at this point so this is what we have left over so now let's replace x with four so it's negative one divided by two square root four times two plus square root four so the square root of four is two and two times two that's going to be four and two plus two is four and four times four is sixteen so the final answer is negative one divided by sixteen and so that's going to be the limit now let's talk about how to evaluate limits graphically so let's say if we want to calculate the limit as x approaches negative 3 from the left side and let's say this graph represents the function f of x so what can we do so to evaluate the limit you're looking for the y value so first identify where x is negative three x is negative 3 anywhere along that vertical line now we want to approach that vertical line from the left side so therefore you want to follow the curve from the left until you get to that point so notice that the y value here corresponds to positive one so therefore the limit as x approaches three from the left side is one now what about from the right side so you want to approach the vertical line at negative 3 from the right so we got to follow this curve so notice that the y value here is negative 3. so therefore the limit as x approaches negative three from the right side that's a negative three now what about the limit as x approaches negative three from either side if the left-sided limit and the right-sided limit are not the same then the limit does not exist so these two are known as one-sided limits now what about f of negative three what is the value of the function when x is negative three to find it identify the closed circle which has a y value of negative three so this is it now let's work on some more problems so what is the limit as x approaches negative two from the left side go ahead and try this one so identify the vertical line at negative two so we want to approach that line from the left side so notice that the y value is negative two so what is the limit as x approaches negative two from the right side so this time we want to approach the vertical line from the right and it points to the same value negative two so therefore the limit as x approaches negative two from either side does exist because these two they match so therefore it's going to be negative two now what is the function value when x is negative 2 so look for the closed circle that's on this vertical line and so that's this point where the y value is positive 2. so as you can see is not very difficult to evaluate limits graphically now let me give you a new set of problems evaluate the limit as x approaches positive one from the left side from the right side and from either side and also find the function value when x is one so x is one anywhere along that vertical line so if we approach it from the left side notice that the y value is positive one there and if we approach it from the right side the y value is three so because these two do not match the limit as x approaches one does not exist now the function value at one is the closed circle which has a y value of two now here's the last set of problems like this what is the limit as x approaches positive 3 from the left side and from the right side and from either side and then find the function value at 3. so notice that at x equals 3 we have this vertical asymptote so as x approaches 3 from the right i mean from the left side notice that it goes down to negative infinity and as we approach 3 from the right side it goes all the way up to positive infinity now these two do not match so therefore the limit does not exist and the function value at 3 is going to be undefined so a good example of having a vertical asymptote at x equals 3 would be a function like this one over x minus three and if you plug in three you're gonna get one over zero which is undefined so whenever you have a zero in the denominator at that point you have a vertical asymptote and it's undefined at that point now at negative 3 we have what is known as a jump discontinuity the graph doesn't connect at negative two we have what is known as a hole a hole is a removable discontinuity a jump discontinuity is not removable so this is another example of a jump discontinuity it's a non-removable discontinuity and here we have an infinite discontinuity which is also non-removable