Limits

The limiting value of a function f(x) at x = a is the value that f(x) approaches to when x approaches to a. This limiting value can be a finite number or ∞ or -∞. Sometimes, it is impossible to determine the limiting value of f(x) when x approaches to a.

To define this limiting value of a function f(x) at x = a, we must consider whether x approaches to a from the left side (x always less than a while approaching to a) or whether x approaches to a from the right side (x always more than a while approaching to a) . These two limiting values of f(x) are called the left and right limits. If these two left and right limits are equal, then only we can say that we can determine the limiting value of f(x) as x approaches to a and define it be equal to the equal left and right limits as follows:

$\lim_{x \to a^{-}} f (x) = \lim_{x \to a^{+}} f (x) = \lim_{x \to a} f (x)$
where the first is the limiting value of the function f(x) when x approaches to a from the left side, the second is the limiting value of f(x) when x approaches to a from the right side, and the third simply the limiting value of f(x) when x approaches to a.

If the left or right limit does not exist, or if both exist but are not equal, then we say the third limit does not exist. It is possible that we simply can not determine the left or right limit, and in such cases we say that the left or right limit do not exist, and of course in that case the third limit or the actual limit of f(x) does not exist as x approaches to a.

We will rarely consider cases when the left or right limit does not exist. You may see such cases mostly as counter examples in advanced mathematics.

In case $a = \infty$ or, $a = - \infty$ , we can only talk about x approaching from one side. For example x can only approach to $\infty$ from the left side and x can only approach to $- \infty$ from the right side. In these cases, the third limit is just the one sided limit. That is:

$\lim_{x \to \infty} f (x) = \lim_{x \to \infty^{-}} f (x) and \lim_{x \to - \infty} f (x) = \lim_{x \to - \infty^{+}} f (x)$
Consider the below function

$f (x) = {\begin{cases} 0, if the greatest integer less than x is even \\ 1, if the greatest integer less than x is odd \end{cases}$
Now, that's weird looking definition of a function. As you can see, if x is between 0 and 1 then f(x) = 0, if x is between 1 and 2 then f(x) is 1, if x is between 2 and 3 then f(x) is 0, so on. So, if we want to find out the limiting value of f(x) when x approaches to ∞, then we have a problem. The function f(x) keeps fluctuating between 0 and 1 and never becomes steady in any large interval. Thus, there is no limiting value of f(x) as x approaches to ∞. In other words, $\lim_{x \to \infty} f (x)$ does not exist. This may be the last time you will see this kind of weird function.

In this course, we will mainly deal with polynomial and rational functions while dealing with limits, and perhaps a few artificial functions.

Polynomials are nicest functions you can think of. They have nice curves. Rational functions are not as great, but they have nice curves too.

The good thing about polynomials is really their curves. They are smooth and very predictable. They don’t swing wildly. This is because the value of a polynomial function p(x) does not change much if you change the value of x just slightly. For example, the value of p(x) at x = 2 is not going to be any big different than the value of p(x) at x = 1.9, or 1.99 or 1.99 or 1.999 or 1.999, … or 2.1 or 2.01, or 2.001 or 2.0001 and so on. This means that

$\lim_{x \to 2^{-}} p (x) = \lim_{x \to 2^{+}} p (x) = p (2)$
And, that’s the case if we replace 2 by any real number. So, if p(x) is a polynomial and you want to calculate $\lim_{x \to a} p (x)$ where a is any real number, then you just evaluate p(a). In other words,
$\lim_{x \to a} p (x) = p (a)$
It’s a little different if a = $\infty$ or a = - $\infty$ . The value of p(x) simply gets too large in size when x gets too large in size. If you think of it, if x is very large in size, then p(x) is basically $k x^{n}$ , where n is the degree of p(x). This is because the smaller powers of x would be just too small compared to the value of $k x^{n}$ and can be just ignored. So, as you can imagine

$\begin{array}{l} \lim_{x \to \infty} p (x) = \infty, if k is positive \\ \lim_{x \to \infty} p (x) = - \infty, if k is negative \\ \lim_{x \to - \infty} p (x) = \infty, if k is positive and n is even \\ \lim_{x \to - \infty} p (x) = - \infty, if k is positive and n is odd \\ \lim_{x \to - \infty} p (x) = \infty, if k is negative and n is odd \\ \lim_{x \to - \infty} p (x) = - \infty, if k is negative and n is even \end{array}$
If you have the polynomial p(x) completely factored and you have the sign graph of p(x) in front of you, then you can tell the limits when x approaches to $\infty$ or - $\infty$ . Just look at the sign of p(x) at the extreme left and extreme right. If the sign on the extreme right is positive, then the $\lim_{x \to \infty} p (x) = \infty$ , or if the sign on the extreme right is negative then $\lim_{x \to \infty} p (x) = - \infty$ , if the sign on the extreme left is positive then $\lim_{x \to - \infty} p (x) = \infty$ , and finally if the sign on the extreme left is negative then $\lim_{x \to - \infty} p (x) = - \infty$ , as simple as that.

As for rational functions, their curves are nice, but not as nice as polynomial curves. In truth, except for the extreme left and right ends, polynomial curves and rational curves look almost the same except near those values of x where the denominator in the rational function becomes zero. Well, naturally. The rational function will not be defined if the denominator were zero for one can not divide by zero. Take a look at the following rational function
$r (x) = \frac{1}{x - 1}$

As long as x is not near 1, the value of r(x) does not change much as we change the value of x. And, even when x is near 1, and x stays either to the left or to the right of 1, the value of r(x) does not change much (comparatively) as we change the value of x. What is happening is that when x is near 1, (x $-$ 1) is very close to zero, and when we divide 1 by a number very close to zero, we get something very large in size. Look at the following table and see the values of r(x) as the value of x approaches to 1 from the right and left..

$x \to 1^{+}$
r(x)
$x \to 1^{-}$
r(x)

1.1
10
.9
-10

1.01
100
.99
-100

1.001
1000
.999
-1000

1.0001
10,000
.9999
-10,000

1.00001
100,000
.99999
-100,000

As you can see, the value of r(x) approaches to $∞$ when x approaches to 1 from the right side and approaches to - $\infty$ when x approaches to 1 from the left side. So,

$\lim_{x \to 1^{+}} \frac{1}{x - 1} = \infty and \lim_{x \to 1^{-}} \frac{1}{x - 1} = - \infty$

And, that's the case always in a rational function when x approaches to a number at which the denominator is zero but the numerator is not. You just have to be careful when the numerator and denominator both become zero for some value of x. In such cases, first thing you want to do is reduce the rational function by canceling out any common factor from the numerator and denominator. We will deal with such cases a little later. Lets just assume that in our rational functions, there is no common factor between the numerator and denominator, meaning both numerator and denominator both are not zero for any value of x.

So, if for some value $x_{0}$ of x, the numerator of the rational function r(x) is not zero but the denominator is, then

$\lim_{x \to x_{0}^{+}} r (x) = \pm \infty and \lim_{x \to x_{0}^{-}} r (x) = \pm \infty$
You can draw the sign graph of r(x) and see what is the sign on the immediate left and right of x = $x_{0}$ , and determine if the above limit is $\infty$ or - $\infty$ . If you don't want to find the sign graph just to find the limit, consider values of x that are sufficiently close to $x_{0}$ and far away from other values of x for which the numerator or denominator become zero, and find the sign of r(x), if it is positive the limit will be positive and if negative the limit will be negative.

If $x_{0}$ is just any other value of x, then $\lim_{x \to x_{0}} r (x) = r (x_{0})$ .

To find $\lim_{x \to \pm \infty} r (x)$ , we consider the degrees of the numerator and denominator.

If the degree of the denominator is more than degree of the numerator, then $\lim_{x \to \pm \infty} r (x)$ = 0, as simple as that. This is because $r (x) ≅ \frac{1}{a x^{k}}$ , where k is the positive difference between the degrees of the numerator and denominator, when x is very large in size ( I will explain this in an example later) and $\frac{1}{a x^{k}}$ approaches to 0 as x approaches to $\pm \infty$ .

If the degree of the numerator is more than the degree of denominator $\lim_{x \to \pm \infty} r (x)$ = $\pm \infty$ . To determine the sign, you may want to refer to the sign graph of r(x) as in the polynomial case. This is because $r (x) ≅ a x^{k}$ , where k is the positive difference between the degrees of the numerator and denominator, when x is very large in size (again, I will explain this a little later in an example), and $a x^{k}$ approaches to $\pm \infty$ as x approaches to $\pm \infty$ as in the case of polynomials.

Lastly, if the degree of the numerator and denominator are the same, then $\lim_{x \to \pm \infty} r (x) = \frac{a}{b}$ . This is because $r (x) ≅ \frac{a}{b}$ , where a and b are the leading coefficients of the numerator and denominator, when x is very large in size. I will now explain each of three cases by way of examples.

Suppose $r (x) = \frac{2 x^{2} - 3 x + 5}{5 x^{2} + 4 x - 3}$ . Note that the degree of the numerator and the denominator are the same, both are 2. Now, if one divides each term in the numerator and denominator by $x^{2}$ (the equal degree), then we get $r (x) = \frac{2 - \frac{3}{x} + \frac{5}{x^{2}}}{5 + \frac{4}{x} - \frac{3}{x^{2}}}$ . As you can see, except for the terms 2 and 5, the rest become very small $-$ in fact very negligible, when x is very large in size. So, $\lim_{x \to \pm \infty} r (x) = \frac{2}{3}$ .

If the degree of the denominator is more, then suppose $r (x) = \frac{2 x^{2} - 3}{4 x^{3} + 7 x^{2} + x - 5}$ . Then divide every term in the denominator and denominator by $x^{2}$ (the smaller degree), and we get $r (x) = \frac{2 - \frac{3}{x^{2}}}{4 x + 7 + \frac{1}{x} - \frac{5}{x^{2}}}$ . Again, as you can see all the terms except 2, 4x and 7 become very small when x becomes very large in size. So, $r (x) ≅ \frac{2}{4 x + 7}$ when x is very large in size and which in turn approaches to 0 when x approaches to $\pm \infty$ . Thus, $\lim_{x \to \pm \infty} \frac{2 x^{2} - 3}{4 x^{3} + 7 x^{2} + x - 5} = 0$
If the degree of the numerator is more, then suppose $r (x) = \frac{x^{2} - 6 x + 2}{x + 4}$ . Then divide all the terms by $x^{2}$ and get $r (x) = \frac{x - 6 + \frac{2}{x}}{1 + \frac{4}{x}} ≅ \frac{x - 6}{1} = (x - 6)$ when x is very large in size, and (x $-$ 6) in turn gets very large in size when x gets very large in size. Thus, $\lim_{x \to \pm \infty} \frac{x^{2} - 6 x + 2}{x + 4} = \pm \infty$

Given the above discussion, you should now be able to find the limit of any polynomial or rational function f(x) when x approaches to some finite number a or to $\pm \infty$ .

Below are some worked out examples:

Example 1

$\begin{array}{l} \lim_{x \to 3} (3 x^{2} - 5 x + 7) \\ = 3 {(3)}^{2} - 5 (3) + 7 \\ = 3 (9) - 15 + 7 \\ = 27 - 15 + 7 \\ = 27 - 8 \\ = 19 \end{array}$

Example 2

$\begin{array}{l} \lim_{x \to - 1} (3 x^{3} - 5 x^{2} + 4 x - 2) \\ = 3 {(- 1)}^{3} - 5 {(- 1)}^{2} + 4 (- 1) - 2 \\ = 3 (- 1) - 5 (1) - 4 - 2 \\ = - 3 - 5 - 4 - 2 \\ = - 14 \end{array}$

Example 3

$\begin{array}{l} \lim_{x \to 0} (3 x^{5} - x + 8) \\ = 3 {(0)}^{5} - (0) + 8 \\ = 3 (0) - 0 + 8 \\ = 0 - 0 + 8 \\ = 8 \end{array}$

Example 4

$\begin{array}{l} \lim_{x \to \infty} (6 x^{5} - 7 x^{4} + 3 x^{3} - 9) \\ = \infty, because the leading coefficient 6 is positive \end{array}$

Example 5

$\begin{array}{l} \lim_{x \to \infty} (- 3 x^{8} + 7 x^{3} - 8 x^{2} + 6 x - 2) \\ = - \infty, because the leading coefficient -3 is negative \end{array}$

Example 6

$\begin{array}{l} \lim_{x \to - \infty} (2 x^{4} - 3 x + 1) \\ = \infty, because leading coefficient 2 is positive and the degree 4 is even \end{array}$

Example 7

$\begin{array}{l} \lim_{x \to - \infty} (- 5 x^{8} + 7 x^{3} + 9 x^{2} - 1) \\ = - \infty, because leading coefficient -5 is negative and the degree 8 is even \end{array}$

Example 8

$\begin{array}{l} \lim_{x \to - \infty} (3 x^{5} + 4 x^{2} - 9 x + 15) \\ = - \infty, because leading coefficient 3 is positive and the degree 5 is odd \end{array}$

Example 9

$\begin{array}{l} \lim_{x \to - \infty} (- x^{7} - 5 x^{5} + 3 x^{4} - 9 x + 5) \\ = \infty, because leading coefficient -1 is negative and degree 7 is odd \end{array}$

Example 10

$\begin{array}{l} \lim_{x \to 2} \frac{(x - 1) (x - 5)}{(x + 3) (x - 4) (x + 7)} \\ = \frac{(2 - 1) (2 - 5)}{(2 + 3) (2 - 4) (2 + 7)} \\ = \frac{(1) (- 3)}{(5) (- 2) (9)} \\ = \frac{1}{30} \end{array}$

Example 11

$\begin{array}{l} \lim_{x \to 3^{-}} \frac{x {(x - 2)}^{2}}{(x + 2) (x + 3) (x - 4)} \\ = \frac{3 {(3 - 2)}^{2}}{(3 + 2) (3 + 3) (3 - 4)} \\ = \frac{3 {(- 1)}^{2}}{(5) (6) (- 1)} \\ = \frac{3 (1)}{- 30} \\ = \frac{- 1}{10} \end{array}$

Example 12

$\begin{array}{l} \lim_{x \to 2^{-}} \frac{3 x + 4}{(x - 3) (x - 2)} \\ = \infty, because denominator beceomes zero at 2 and the function is positive to the left of 2 \end{array}$

Example 13

$\begin{array}{l} \lim_{x \to 4^{+}} \frac{(x + 3) {(x - 6)}^{2}}{x (x - 1) (x - 4)} \\ = \infty, because denominator is 0 at 4 and the sign of the function is positive to the right of 4 \end{array}$

Example 14

$\begin{array}{l} \lim_{x \to 0^{+}} \frac{(x + 3) (x - 1)}{x^{2} (x + 2)} \\ = \infty, because denominator is 0 at 0 and the function is positive on the right of 0 \end{array}$

Example 15

$\begin{array}{l} \lim_{x \to 2} \frac{{(x - 1)}^{3} (x + 4)}{{(x - 2)}^{2} (x - 3)} \\ = - \infty, because the denominator is 0 at 2 and the function is negative on both sides of 2 \end{array}$

Example 16

$\begin{array}{l} \lim_{x \to - 1} \frac{x (x + 5) (x - 2)}{{(x + 1)}^{2} (x - 3)} \\ = - \infty, because the function has negative sign on both sides of -1 \end{array}$

Example 17

$\begin{array}{l} \lim_{x \to \infty} \frac{x^{2} + 3 x - 1}{2 x^{2} - 5 x + 2} \\ = \lim_{x \to \infty} \frac{1 + \frac{3}{x} - \frac{1}{x^{2}}}{2 - \frac{5}{x} + \frac{2}{x^{2}}} \\ = \frac{1}{2} \end{array}$

Example 18

$\begin{array}{l} \lim_{x \to \infty} \frac{7 x + 2}{x - 4} \\ = \lim_{x \to \infty} \frac{7 + \frac{2}{x}}{1 - \frac{4}{x}} \\ = \frac{7}{1} = 7 \end{array}$

Example 19

$\begin{array}{l} \lim_{x \to - \infty} \frac{- 3 x^{3} + 4 x^{2} - 5}{5 x^{3} + x^{2} + 7 x - 8} \\ = \lim_{x \to - \infty} \frac{- 3 + \frac{4}{x^{2}} - \frac{5}{x^{3}}}{5 + \frac{1}{x} + \frac{7}{x^{2}} - \frac{8}{x^{3}}} \\ = \frac{- 3}{5} \end{array}$

Example 20

$\begin{array}{l} \lim_{x \to - \infty} \frac{2 x^{4} - 4 x + 3}{- 6 x^{4} + 7 x^{3} - x} \\ = \lim_{x \to - \infty} \frac{2 - \frac{4}{x^{3}} + \frac{3}{x^{4}}}{- 6 + \frac{7}{x} - \frac{1}{x^{3}}} \\ = \frac{2}{- 6} \\ = \frac{- 1}{3} \end{array}$

$x \to 1^{+}$	r(x)	$x \to 1^{-}$	r(x)
1.1	10	.9	-10
1.01	100	.99	-100
1.001	1000	.999	-1000
1.0001	10,000	.9999	-10,000
1.00001	100,000	.99999	-100,000