Understanding Limits in Calculus

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 2 of the function x squared minus 4 divided by x minus 2. So how can we do so? Well one way is to use direct substitution. If we plug in 2, notice what will happen. 2 squared is 4, 4 minus 4 is 0. So 0 over 0 is undefined, which we don't know what value that represents. Now sometimes you could find a limit by plugging a value. that's close to 2. And that's what you want to do. You want to plug in a number that's close to 2 but not exactly 2. 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. 2.1 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 parentheses, you should get 4.01. So notice what's happening. As we get closer and closer to 2, the limit approaches 4. So we could therefore say that the limit as x approaches 2 of this function is equal to 4. 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. Sometimes, you have to use other techniques to get the answer. In this particular example, we could factor. x squared minus 4, you can write it as x plus 2 times x minus 2. Now we need to rewrite the limit expression until we replace x with 2. Now notice that we can cancel x minus 2. So when this term is gone, we can now use direct substitution because the x minus 2 factor was given. us a 0 in the denominator, which we don't want. So now all we need to do is find the limit as x approaches 2 of x plus 2. So now we can replace x with 2, and 2 plus 2 is 4. 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 2x minus 4. So notice that we don't have a fraction. We're not going to get a zero in a denominator. So for a question like this, you can use direct substitution. So all you got to do is plug in 5. So it's going to be 5 squared plus 2 times 5 minus 4. So that's 25 plus 10 minus 4. And 25 plus 10 is 35. So the limit... 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. How can we factor x? cube minus 27. So what we have is the difference of cubes and whenever you see that you can use this formula. a cube minus b cube 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 3. So this is going to be x minus 3, and then a squared, that's x squared, and then plus ab, 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 3. So what we have left over is the limit as x approaches 3 of x squared plus 3x plus 9. So at this point, we now can use... direct substitution. So it's 3 squared plus 3 times 3 plus 9, which is 9 plus 9 plus 9. Adding 9 three times is basically multiplying 9 by 3. And so this limit is equal to 27. Now here's another problem that you can work on. So what is the limit as x approaches 3 of 1 over x minus 1 over 3 divided by x minus 3? 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 3's 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 1. If we take out a negative 1, the negative x will change to positive x, and positive 3 will change to negative 3. 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 3x. Well now we can use direct substitution. So let's replace x with 3. So it's going to be 3 times 3, which is 9. So the final answer is negative 1 divided by 9. 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 9 of square root x minus 3 over x minus 9? 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 a 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 3, so that's going to be plus. 3 square root x, and then these two will form negative 3 square root x. And finally, we have negative 3 times positive 3, which is negative 9. Now on the bottom, I'm not going to FOIL the two expressions. I'm going to leave it. the way it is because my goal is to get rid of the x minus 9. I want to cancel it. So now negative 3 and positive 3 add up to 0. 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 9. And so what we have left over is the limit as x approaches 9 of 1 over square root x plus 3. So now what we can do is replace x with 9. 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 3 is 6. So the final answer is 1 over 6. Now let's look at this example. What is the limit as x approaches 4 of the expression 1 over square root x minus 1 over 2 divided by x minus 4? So this... Next 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 the 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 2. So I have the limit as x approaches 4 with a 2 on top. If I multiply these two, the 2's 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 1. So this is now the limit as x approaches 4. So negative 1, 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 4 terms at this point. So this is what we have left over. So now let's replace x with 4. So it's negative 1 divided by 2 square root 4 times 2 plus square root 4. So the square root of 4 is 2. And 2 times 2, that's going to be 4. And 2 plus 2 is 4. And 4 times 4 is 16. So the final answer is negative 1 divided by 16. 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 3. 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 1. So therefore, the limit as x approaches 3 from the left side is 1. Now what about from the right side? So you want to approach the vertical line at negative 3 from the right. So we've got to follow this curve. So notice that the y value here is negative 3. So therefore, the limit as x approaches negative 3 from the right side That's negative 3. Now what about the limit as x approaches negative 3 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 3? What is the value of the function when x is negative 3? To find it, identify the closed circle, which has a y value of negative 3. So this is it. Now let's work on some more problems. So what is the limit as x approaches negative 2 from the left side? Go ahead and try this one. So identify the vertical line at negative 2. So we want to approach that line from the left side. So notice that the y value is negative 2. So what is the limit as X approaches negative 2 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 2. So therefore, the limit as x approaches negative 2 from either side does exist because these two, they match. So therefore, it's going to be negative 2. 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, it's not very difficult to evaluate limits graphically. Now let me give you a new set of problems. Evaluate the limit as x approaches positive 1 from the left side, from the right side, and from either side, and also find the function value when x is 1. So, x is 1 anywhere along that vertical line. So, if we approach it from the left side, notice that the y value is positive 1 there. And if we approach it from the right side, the y value is 3. So, because these two do not match, the limit as x approaches 1 does not exist. Now the function value at 1 is the closed circle which has a y value of 2. 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 left side, notice that it goes down to... negative infinity and as we approach tree 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 three is going to be undefined So a good example of having a vertical asymptote at x equals 3 would be a function like this, 1 over x minus 3. And if you plug in 3, you're going to get 1 over 0, which is undefined. So whenever you have a 0 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 2, 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.

Well one way is to use direct substitution. If we plug in 2, notice what will happen. 2 squared is 4, 4 minus 4 is 0. So 0 over 0 is undefined, which we don't know what value that represents.

Now sometimes you could find a limit by plugging a value. that's close to 2. And that's what you want to do. You want to plug in a number that's close to 2 but not exactly 2. 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.

2.1 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 parentheses, you should get 4.01. So notice what's happening. As we get closer and closer to 2, the limit approaches 4. So we could therefore say that the limit as x approaches 2 of this function is equal to 4. 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. Sometimes, you have to use other techniques to get the answer.

In this particular example, we could factor. x squared minus 4, you can write it as x plus 2 times x minus 2. Now we need to rewrite the limit expression until we replace x with 2. Now notice that we can cancel x minus 2. So when this term is gone, we can now use direct substitution because the x minus 2 factor was given. us a 0 in the denominator, which we don't want. So now all we need to do is find the limit as x approaches 2 of x plus 2. So now we can replace x with 2, and 2 plus 2 is 4. 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 2x minus 4. So notice that we don't have a fraction. We're not going to get a zero in a denominator.

So for a question like this, you can use direct substitution. So all you got to do is plug in 5. So it's going to be 5 squared plus 2 times 5 minus 4. So that's 25 plus 10 minus 4. And 25 plus 10 is 35. So the limit... 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. How can we factor x? cube minus 27. So what we have is the difference of cubes and whenever you see that you can use this formula.

a cube minus b cube 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 3. So this is going to be x minus 3, and then a squared, that's x squared, and then plus ab, 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 3. So what we have left over is the limit as x approaches 3 of x squared plus 3x plus 9. So at this point, we now can use... direct substitution.

So it's 3 squared plus 3 times 3 plus 9, which is 9 plus 9 plus 9. Adding 9 three times is basically multiplying 9 by 3. And so this limit is equal to 27. Now here's another problem that you can work on. So what is the limit as x approaches 3 of 1 over x minus 1 over 3 divided by x minus 3? 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 3's 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 1. If we take out a negative 1, the negative x will change to positive x, and positive 3 will change to negative 3. 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 3x. Well now we can use direct substitution.

So let's replace x with 3. So it's going to be 3 times 3, which is 9. So the final answer is negative 1 divided by 9. 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 9 of square root x minus 3 over x minus 9?

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 a 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 3, so that's going to be plus. 3 square root x, and then these two will form negative 3 square root x. And finally, we have negative 3 times positive 3, which is negative 9. Now on the bottom, I'm not going to FOIL the two expressions.

I'm going to leave it. the way it is because my goal is to get rid of the x minus 9. I want to cancel it. So now negative 3 and positive 3 add up to 0. 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 9. And so what we have left over is the limit as x approaches 9 of 1 over square root x plus 3. So now what we can do is replace x with 9. 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 3 is 6. So the final answer is 1 over 6. Now let's look at this example. What is the limit as x approaches 4 of the expression 1 over square root x minus 1 over 2 divided by x minus 4? So this...

Next 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 the 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 2. So I have the limit as x approaches 4 with a 2 on top. If I multiply these two, the 2's 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 1. So this is now the limit as x approaches 4. So negative 1, 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 4 terms at this point.

So this is what we have left over. So now let's replace x with 4. So it's negative 1 divided by 2 square root 4 times 2 plus square root 4. So the square root of 4 is 2. And 2 times 2, that's going to be 4. And 2 plus 2 is 4. And 4 times 4 is 16. So the final answer is negative 1 divided by 16. 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 3. 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 1. So therefore, the limit as x approaches 3 from the left side is 1. Now what about from the right side? So you want to approach the vertical line at negative 3 from the right.

So we've got to follow this curve. So notice that the y value here is negative 3. So therefore, the limit as x approaches negative 3 from the right side That's negative 3. Now what about the limit as x approaches negative 3 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 3? What is the value of the function when x is negative 3? To find it, identify the closed circle, which has a y value of negative 3. So this is it. Now let's work on some more problems.

So what is the limit as x approaches negative 2 from the left side? Go ahead and try this one. So identify the vertical line at negative 2. So we want to approach that line from the left side.

So notice that the y value is negative 2. So what is the limit as X approaches negative 2 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 2. So therefore, the limit as x approaches negative 2 from either side does exist because these two, they match. So therefore, it's going to be negative 2. 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, it's not very difficult to evaluate limits graphically. Now let me give you a new set of problems. Evaluate the limit as x approaches positive 1 from the left side, from the right side, and from either side, and also find the function value when x is 1. So, x is 1 anywhere along that vertical line. So, if we approach it from the left side, notice that the y value is positive 1 there. And if we approach it from the right side, the y value is 3. So, because these two do not match, the limit as x approaches 1 does not exist.

Now the function value at 1 is the closed circle which has a y value of 2. 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 left side, notice that it goes down to... negative infinity and as we approach tree 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 three is going to be undefined So a good example of having a vertical asymptote at x equals 3 would be a function like this, 1 over x minus 3. And if you plug in 3, you're going to get 1 over 0, which is undefined. So whenever you have a 0 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 2, 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.

Transcript for:Understanding Limits in Calculus

Transcript for:
Understanding Limits in Calculus