 So in calculus, the concept of infinity becomes important. And one place it emerges is in the following. Suppose I want to find the limit as x approaches 0 from above of 1 over x. So I'll try out some x values that are close to 0, but always staying a little bit above it. 0.1, 0.01, 0.001, and I'll evaluate 1 over x. Now there is no number that these values approach, so there is no limit in the usual sense. However, these values have the property of getting larger and larger without bound. To record this observation, we say that the limit is infinity, and we record that using our infinity symbol. Sometimes we'll write this as plus infinity for specificity, because these are large positive numbers. And it is vitally important to remember that the symbol of infinity expresses the idea of a quantity that increases without bound. It is not a number. Similarly, we might ask what happens to the limit as x approaches 0 from below of 1 over x. So I'll take a look at x values that are close to but less than 0, negative 0.1, negative 0.01, negative 0.001, and I'll find 1 over x is minus 10, minus 100, minus 1000, and our values are becoming more and more negative without bound. And to record this observation, we say the limit is negative infinity, and write it using a negative infinity notation. If our infinity symbol indicates something that is becoming more and more, whether positive or negative, without bound, then we can also talk about the limit as x goes to infinity of a function. For example, let's look at the limit as x goes to infinity of 1 over x. So as before, we'll consider x values that are getting larger and larger and larger, and we'll calculate our 1 over x values. And here, these values do seem to be approaching a number. We round these to the same number of decimal places we get 0. And so this table suggests that the limit as x goes to infinity of 1 over x is 0. We can go a little bit further. Nothing significant would change if our denominator was x squared, or x cubed, or square root of x, or in general x raised to any positive power. And so this suggests the following theorem. If our exponent n is greater than 0, then the limit as x goes to infinity of 1 over x to the n is equal to 0. And if we combine this with all of our other limit theorems, then we'll be able to calculate many limits as x goes to infinity. For example, let's find this limit, and it will always be useful to determine what type of function this is. So remember, the type of function is determined by the last operation performed, and in this case, we're adding, and so this is a sum. So we can use our theorem about the limit of a sum is the sum of the limits. Now if we take a look at the individual functions here, this 3 over x and this 5 over x squared, we can rewrite those as 3 times 1 over x and 5 times 1 over x squared. And now these functions are constant multiples, and so we can invoke our constant multiple theorem. And now we're trying to find the limit as x goes to infinity of 1 over x and the limit as x goes to infinity of 1 over x squared, but we have a theorem that tells us the limit as x goes to infinity of 1 over x to the n. And that allows us to find our limit. As another example, we might find the limit as x goes to infinity of the square root of 1 plus 4 over x squared. And the type of function is determined by the last operation, and so this is a root and the limit of a root is the root of the limit and the limit of a sum is the sum of the limits and we can evaluate these limits directly. And this will give us a couple more indeterminate and determinant forms. Suppose the limit as x approaches a if some function is infinity, the limit of g is also infinity and the limit of h is l, some determinant number. Well, the limit of h of x over f of x, that's the limit of a quotient, is the quotient of the limits will be l over infinity. This is going to be determinant with the value of 0 and that's because I'm taking a number and dividing it by an increasingly larger and larger number. Likewise, I can take the limit as x approaches a of f of x over h of x. Again, this is the limit of a quotient, so that will be the quotient of the limits and this will have form infinity over l and this is determinant with the value of infinity. And then finally, the limit as x approaches a of f of x over g of x is indeterminate because it has form infinity over infinity and as with other indeterminate forms, we can often simplify these algebraically. So let's try and find the limit of a rational function. Since it's a quotient, we can find the limits individually, however, both numerator and denominator do have a limit of infinity, so this is an indeterminate form. Now, if we could somehow change this so our denominator has a non-zero limit, we can find the limit of the quotient and this leads to the following idea. If we have a rational function, we'll multiply the numerator and denominator by 1 over x to the n, where x to the n is the highest power of x in the denominator. So in this rational function, the highest power of x in the denominator is x squared and so we'll multiply numerator and denominator by 1 over x squared. And we'll do some algebraic simplifications and since this is the limit of a quotient, we can evaluate this as the quotient of the limits and get a limit value of zero. Now, let's also find some support for this answer numerically. So we're trying to find out what happens to this expression as x gets larger and larger, so we'll pick some large values for x at x equals 100 or 1000 or 10000. It appears we are getting values close to zero, which supports our claimed limit of zero. What about a more complex expression? So the highest power of x in the denominator appears to be x squared, but because it's under a radical, it's really the square root of x squared, which reduces down to x, since if x goes to infinity, if x gets larger and larger, x will be a positive number. So we'll multiply numerator and denominator by 1 over the square root of x squared, otherwise known as 1 over x, and then do some algebra to simplify the expression. And we'll apply our limit theorems to find our limit, which will be 3 over 1 or 3.