Calculus/Limits

Intuitive Look

A limit looks at what happens to a function when the input approaches a certain value. The general notation for a limit is as follows:

$\underset{x\to a}{\lim }f(x)=L$

This is read as "The limit of f(x) as x approaches a is L". How we can determine whether a limit exists for f(x) at a and what the limit is is a technical point that we'll take up later. For now, we'll look at it from an intuitive standpoint.

Let's say that the function that we're interested in is $f(x)=x^2$, and that we're interested in its limit as x approaches 2. We can write the limit that we're interested in using the above notation as follows:

$\underset{x\to 2}{\lim }{x}^2$

One way to try to evaluate what this limit is would be to choose values near 2, compute f(x) for each, and see what happens as they get closer to 2. This is implemented as follows:

$x$	1.7	1.8	1.9	1.95	1.99	1.999
$f(x)=x^2$	2.89	3.24	3.61	3.8025	3.9601	3.996001

Here we chose numbers smaller than 2, and approached 2 from below. We can also choose numbers larger than 2, and approach 2 from above:

$x$	2.3	2.2	2.1	2.05	2.01	2.001
$f(x)=x^2$	5.29	4.84	4.41	4.2025	4.0401	4.004001

We can see from the tables that as x grows closer and closer to 2, f(x) seems to get closer and closer to four, regardless of whether we approach 2 with x from above or from below. For this reason, we feel reasonably confident that the limit of $x^2$ as x approaches 2 is 4, or written in limit notation,

$\underset{x\to 2}{\lim }{x}^2=4$

Now let's look at another example. Suppose we're interested in the function $f(x)=\frac{1}{x-2}$, and its behavior as x approaches 2. Here's the limit in limit notation:

$\underset{x\to 2}{\lim }\frac{1}{x-2}$

Just as before, we can compute function values as x approaches 2 from below and from above. Here's a table, approaching from below:

$x$	1.7	1.8	1.9	1.95	1.99	1.999
$f(x)=\frac{1}{x-2}$	-3.333	-5	-10	-20	-100	-1000

And here from above:

$x$	2.3	2.2	2.1	2.05	2.01	2.001
$f(x)=\frac{1}{x-2}$	3.333	5	10	20	100	1000

In this case, the function doesn't seem to be approaching any value as x approaches 2. In this case we would say that the limit doesn't exist.

Both of these examples may seem trivial, but consider the following function:

$f(x)=\{\begin{array}{cc}{x}^2& \text{if }x\ne 2\\ \text{undefined}& \text{if }x=2\end{array}$ OR $\begin{array}{ccc}f(x)& =& \frac{x^2(x-2)}{x-2}\end{array}$

Now, we see that these functions are completely identical; not just "almost the same," but actually, in terms of the definition of a function, completely the same. However, we are more comfortable with the second one algebraically, because it is easier to work with.

In algebra, we would simply say that we can cancel the term $(x-2)$, and then we have the function $f(x)=x^2$. This, however, would be a bit dishonest; the function that we have now is not really the same as the one we started with, because it is defined at x=2, and our original function was, specifically, not defined at x=2. In algebra we were willing to ignore this difficulty because we had no better way of dealing with this type of function. Now, however, in calculus, we can introduce a better, more correct way of looking at this type of function. What we want is to be able to say that, even though at $x=2$ the function doesn't exist, it works almost as though it does, and it's 4. It may not get there, but it gets really, really close. The only question that we have is: what do we mean by close?

Informal definition of a limit

As the precise definition of a limit is a bit technical, it is easier to start with an informal definition, and we'll explain the formal definition later.

We suppose that a function f is defined for x near c (but we do not require that it be defined when x=c).

Definition: (Informal definition of a limit) We call L the limit of f(x) as x approaches c , i.e. f(x) becomes close to L when x is close (but not equal) to c. When this holds we write $\underset{x\to c}{\lim }f(x)=L$ or $f(x)\to L\text{as}x\to c.$

Notice that the definition of a limit is not concerned with the value of f(x) when x=c (which may exist or may not). All we care about is the values of f(x) when x is close to c, on either the left or the right (i.e. less or greater).

Limit rules

Now that we have defined, informally, what a limit is, we will list some rules that are useful for manipulating a limit. These will all be proven, or left as exercises, once we formally define the fundamental concept of the limit of a function.

Suppose that $\underset{x\to c}{\lim }f(x)=L$ and $\underset{x\to c}{\lim }g(x)=M$ and that k is constant. Then

$\begin{array}{ccccc}\underset{x\to c}{\lim }kf(x)& =& k\cdot \underset{x\to c}{\lim }f(x)& =& kL\\ \underset{x\to c}{\lim }[f(x)+g(x)]& =& \underset{x\to c}{\lim }f(x)+\underset{x\to c}{\lim }g(x)& =& L+M\\ \underset{x\to c}{\lim }[f(x)-g(x)]& =& \underset{x\to c}{\lim }f(x)-\underset{x\to c}{\lim }g(x)& =& L-M\\ \underset{x\to c}{\lim }[f(x)\cdot g(x)]& =& \underset{x\to c}{\lim }f(x)\cdot \underset{x\to c}{\lim }g(x)& =& L\cdot M\\ \underset{x\to c}{\lim }\frac{f(x)}{g(x)}& =& \frac{\underset{x\to c}{\lim }f(x)}{\underset{x\to c}{\lim }g(x)}& =& \frac{L}{M}& \text{as long as }M\ne 0\end{array}$

Notice that in the last rule we need to require that M is not equal to zero (otherwise we would be dividing by zero which is an undefined operation).

These rules are known as identities; they are scalar multiplication, addition, subtraction, multiplication, and division of limits. (A scalar is a constant, and we say that when you multiply a function by a constant, you are performing scalar multiplication.)

Using these rules we can deduce some others. First, the constant rule states that if f(x) = b is constant for all x then the limit as x approaches c must be equal to b. In other words

$\underset{x\to c}{\lim }b=b$.

Second, the identity rule states that if f(x) = x then the limit of f as x approaches c is equal to c. That is,

$\underset{x\to c}{\lim }x=c$.

Third, using the rule for products many times we get that

$\underset{x\to c}{\lim }f(x{)}^n={(\underset{x\to c}{\lim }f(x))}^n$ for a positive integer n.

This is called the power rule.

Examples

Example 1 Find the limit $\underset{x\to 2}{\lim }4x^3.$

We need to simplify the problem, since we have no rules about this expression by itself. We know from the identity rule above that $\underset{x\to 2}{\lim }x=2$. By the power rule, $\underset{x\to 2}{\lim }{x}^3=2^3=8$. Lastly, by the scalar multiplication rule, we get $\underset{x\to 2}{\lim }4x^3=4\cdot 8=32$

Example 2

Find the limit $\underset{x\to 2}{\lim }4x^3+5x+7.$

To do this informally, we split up the expression, once again, into its components. As above,$\underset{x\to 2}{\lim }4x^3=32$.

Also $\underset{x\to 2}{\lim }5x=5\cdot \underset{x\to 2}{\lim }x=5\cdot 2=10$

and $\underset{x\to 2}{\lim }7=7$. Adding these together gives

$\underset{x\to 2}{\lim }4x^3+5x+7=32+10+7=49.$

Example 3

Find the limit: $\underset{x\to 2}{\lim }$ $\frac{(4x^3+5x+7)}{(x-4)(x+10).}$

From the previous example the limit of the numerator is $\underset{x\to 2}{\lim }4x^3+5x+7=49.$ The limit of the denominator is

$\underset{x\to 2}{\lim }(x-4)(x+10)=\underset{x\to 2}{\lim }(x-4)\cdot \underset{x\to 2}{\lim }(x+10)=(2-4)\cdot (2+10)=-24.$

As the limit of the denominator is not equal to zero we can divide which gives

$\underset{x\to 2}{\lim }$ $\frac{(4x^3+5x+7)}{(x-4)(x+10)}$ $=-\frac{49}{24}$.

Example 4

Find the limit: $\underset{x\to 4}{\lim }$ $\frac{(x^4-16x+7)}{(4x-5).}$

We apply the same process here as we did in the previous set of examples;

$\underset{x\to 4}{\lim }$ $\frac{(x^4-16x+7)}{(4x-5)}$ $=$ $\frac{\underset{x\to 4}{\lim }(x^4-16x+7)}{\underset{x\to 4}{\lim }(4x-5)}$ $=$ $\frac{(\underset{x\to 4}{\lim }(x^4)-\underset{x\to 4}{\lim }(16x)+\underset{x\to 4}{\lim }(7))}{\underset{x\to 4}{\lim }(4x)-\underset{x\to 4}{\lim }(5)}$

We can evaluate each of these; $\underset{x\to 4}{\lim }(x^4)=256.$ $\underset{x\to 4}{\lim }(16x)=64.$ $\underset{x\to 4}{\lim }(7)=7.$ $\underset{x\to 4}{\lim }(4x)=16.$ $\underset{x\to 4}{\lim }(5)=5.$ $=\frac{199}{11}$

Example 5

Evaluate the limit $\underset{x\to 0}{\lim }$ $\frac{1-\cos (x)}{x}$.

To evaluate this seemingly complex limit, be aware of your sine and cosine identities. We will also have to use two new facts. First, if f(x) is a trigonometric function (that is, one of sine, cosine, tangent, cotangent, secant or cosecant) and is defined at a, then $\underset{x\to a}{\lim }f(x)=f(a)$. Second, $\underset{x\to 0}{\lim }$ $\frac{\sin (x)}{(x)}$ $=1$.

To evaluate the limit, recognize that $1-\cos (x)$ can be multiplied by $1+cos(x)$ to obtain $(1-co{s}^2(x))$ which, by our trig identities, is $si{n}^2(x)$. So, multiply the top and bottom by $1+cos(x)$ (This is allowed because it is identical to multiplying by one). This is a standard trick for evaluating limits of fractions; multiply the numerator and the denominator by a carefully chosen expression which will make the expression simplify somehow. In this case, we should end up with:

$\underset{x\to 0}{\lim }$ $\frac{1-\cos (x)}{x}$ $=$ $\frac{1-\cos (x)}{x}$ $\cdot$ $\frac{1}{1}$ $=$ $\frac{1-\cos (x)}{x}$ $\cdot$ $\frac{1+\cos (x)}{1+\cos (x)}$ $=$ $\frac{[(1-\cos (x))\cdot 1]+[(1-\cos (x))\cdot \cos (x)]}{x\cdot (1+\cos (x))}$ $=$ $\frac{1-\cos (x)+\cos (x)-{\cos }^2(x)}{x\cdot (1+\cos (x))}$ $=$ $\frac{1-{\cos }^2(x)}{x\cdot (1+\cos (x))}$ $=$ $\frac{{\sin }^2(x)}{x\cdot (1+\cos (x))}$ $=$ $\frac{\sin (x)\cdot \sin (x)}{x\cdot (1+\cos (x))}$ $=$ $\frac{\sin (x)}{x}$ $\cdot$ $\frac{\sin (x)}{(1+\cos (x))}$

Your next step shall be to break this up into $\underset{x\to 0}{\lim }$$\frac{\sin (x)}{(x)}$$\cdot$$\underset{x\to 0}{\lim }$$\frac{\sin (x)}{(1+\cos (x))}$ by the limit rule of multiplication. By the fact mentioned above, $\underset{x\to 0}{\lim }$$\frac{\sin (x)}{(x)}$$=1$.

Next, $\underset{x\to 0}{\lim }$$\frac{\sin (x)}{(1+\cos (x))}$ $=$ $\frac{\underset{x\to 0}{\lim }\sin x}{\underset{x\to 0}{\lim }(1+\cos x)}$ $=$ $\frac{0}{1+\cos 0}$ $=0$.

Thus, by multiplying these two results, we obtain 0.

We will now present an amazingly useful result, even though we cannot prove this yet. We can find the limit of any polynomial or rational function, as above, as long as that rational function is defined at c (so we are not dividing by zero). More precisely, c must be in the domain of the function.

Limits of Polynomials and Rational functions If f is a polynomial or rational function that is defined at c then $\underset{x\to c}{\lim }f(x)=f(c)$

We already learned this for trigonometric functions, so we see that it is easy to find limits of polynomial, rational or trigonometric functions wherever they are defined. In fact, this is true even for combinations of these functions; thus, for example, $\underset{x\to 1}{\lim }(\sin (x^2)+4{\cos }^3(3x-1))=\sin 1^2+4{\cos }^3(3(1)-1)$.

Finding limits

Now, we will discuss how, in practice, to find limits. First, if the function can be built out of rational, trigonometric, logarithmic and exponential functions, then if a number c is in the domain of the function, then the limit is simply the value of the function at c.

If c is not in the domain of the function, then in many cases (as with rational functions) the domain of the function includes all the points near c, but not c. An example would be if we wanted to find $\underset{x\to 0}{\lim }\frac{x}{x}$, where the domain includes all numbers besides 0. In that case, we want to find a similar function, except with the hole filled in. The limit of this function at c will be the same, as can be seen from the definition of a limit. The function is the same as the previous except at a point c. The limit definition depends on f(x) only at the points where x is close to c but not equal to it. When $x\ne c$, this condition doesn't hold, and so the limit at c does not depend on the value of the function at c. Therefore, the limit of the new function is the same as of the previous function. And since the domain of our new function includes c, we can now (assuming it's still built out of rational, trigonometric, logarithmic and exponential functions) just evaluate the function at c as before.

In our example, this is easy; canceling the xs gives 1, which equals x/x at all points except 0. Thus, we have $\underset{x\to 0}{\lim }\frac{x}{x}=\underset{x\to 0}{\lim }1=1$. In general, when computing limits of rational functions, it's a good idea to look for common factors in the numerator and denominator.

Lastly, note that the limit might not exist at all. There are a number of ways in which this can occur:

"Gap": There is a gap (more than a point wide) in the function where the function is not defined. As an example, in

$f(x)=\sqrt{x^2-16}$

f(x) does not have any limit when -4 ≤ x ≤ 4. There is no way to "approach" the middle of the graph. Note also that the function also has no limit at the endpoints of the two curves generated (at x = -4 and x = 4). For the limit to exist, the point must be approachable from both the left and the right. Note also that there is no limit at a totally isolated point on the graph.

"Jump": If the graph suddenly jumps to a different level, there is no limit. This is illustrated in the floor function (in which the output value is the greatest integer not greater than the input value).

Asymptote: In

$f(x)=\frac{1}{x^2}$

the graph gets arbitrarily high as it approaches 0, so there is no finite limit. In this case we say the limit is infinite.

Infinite oscillation: These next two can be tricky to visualize. In this one, we mean that a graph continually rises above and below a horizontal line. In fact, it does this infinitely often as you approach a certain x-value. This often means that there is no limit, as the graph never approaches a particular value. However, if the height (and depth) of each oscillation diminishes as the graph approaches the x-value, so that the oscillations get arbitrarily smaller, then there might actually be a limit.

The use of oscillation naturally calls to mind the trigonometric functions. An example of a trigonometric function that does not have a limit as x approaches 0 is

$f(x)=\sin \frac{1}{x}.$

As x gets closer to 0 the function keeps oscillating between -1 and 1. It is also true that sin(1/x) oscillates an infinite number of times on the interval between 0 and any positive value of x. The sine function, sin(x), is equal to zero whenever x=kπ, where k is a positive integer. Between each value of k, sin(x) oscillates between 0 and -1 or 0 and 1. Hence, sin(1/x)=0 for every x=1/(kπ). In between consecutive pairs of these values, 1/(kπ} and 1/[(k+1)π], sin(1/x) oscillates from 0, to -1, to 1 and back to 0 and so on. We may also observe that there are an infinite number of such pairs, and they are all between 0 and 1/π. There are a finite number of such pairs between any positive value of x and 1/π which implies that there must be infinitely many between x and 0. From our reasoning we may conclude that as x approaches 0 from the right, the function sin(1/x) does not approach any value. We say that the limit as x approaches 0 from the right does not exist. (In contrast, in the case of a "jump", the limits from each side did exist. Since they weren't equal, though, the regular limit didn't exist.)

Incomplete graph: Let us consider two examples.

First, let f be the constant function f(q)=2 where we specify that q is only allowed to be a rational number. Let $q_0$ be some rational number. Then $\underset{q\to q_0}{\lim }f(q)=2$, since, for q close to $q_0$, f(q) is close to (in fact equals) 2.

Now let g be the similar-looking function defined on the entire real line, but we change the value of the function based on whether x is rational or not.

$g(x)=\{\begin{array}{cc}2,& \text{if }x\text{ is rational}\\ 0,& \text{if }x\text{ is irrational}\end{array}$

Now g has a limit nowhere! For let x be a real number; we show that g can't have a limit at x. No matter how close we get to x, there will be rational numbers (where g will be 2) and irrational numbers (where g will be 0). Thus g has no limit at any real number!

Using limit notation to describe asymptotes

Now consider the function

$g(x)=\frac{1}{x^2}.$

What is the limit as x approaches zero? The value of g(0) does not exist; the value of g(0) is not defined.

$g(0)=\frac{1}{0^2}$

Notice that we can make the function g(x) as large as we like, by choosing a small x, as long as x ≠ 0. For example, to make g(x) equal to one trillion, we choose x to be 10^-6. In this case, we say we can make g(x) arbitrarily large (as large as we like) by taking x to be sufficiently close to zero, but not equal to zero. And we express it algebraically as follows

$\underset{x\to 0}{\lim }g(x)=\underset{x\to 0}{\lim }\frac{1}{x^2}=\mathrm{\infty }$

Note that the limit does not exist at x = 0.

Example: Finding Vertical and Horizontal Asymptotes

Use limits to find all vertical and horizontal asymptotes of the function $f(x)=\frac{1}{x+2}$

The graph of $f(x)=\frac{1}{x+2}$ has a vertical asymptote if f(x) goes to infinity. Therefore, we can determine any vertical asymptotes by setting the denominator equal to 0.

$x+2=0$

$x+2-2=0-2$

$x=-2$

Notice that $f(-2)=\frac{1}{0}$. Therefore, f(x) has a vertical asymptote at x = -2.

The graph of $f(x)=\frac{1}{x+2}$ has a horizontal asymptote if f(x) converges on a value at infinity. We can find any horizontal asymptotes by finding $\underset{x\to \mathrm{\infty }}{\lim }f(x).$

$\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{x+2}=\frac{1}{\mathrm{\infty }+2}=\frac{1}{\mathrm{\infty }}=0$

Therefore f(x) has a horizontal asymptote at y = 0.

Application to calculus

To see the power of the limit, let's consider a moving car. Suppose we have a car whose position is linear with respect to time (that is, a graph plotting the position with respect to time will show a straight line). We want to find the velocity. This is easy to do from algebra, we just take the slope, and that's our velocity.

But unfortunately (or perhaps fortunately if you are a calculus teacher), things in the real world don't always travel in nice straight lines. Cars speed up, slow down, and generally behave in ways that make it difficult to calculate their velocities.

Now what we really want to do is to find the velocity at a given moment (the instantaneous velocity). The trouble is that in order to find the velocity we need two points, while at any given time, we only have one point. We can, of course, always find the average speed of the car, given two points in time, but we want to find the speed of the car at one precise moment.

Here is where the basic trick of differential calculus comes in. We take the average speed at two moments in time, and then make those two moments in time closer and closer together. We then see what the limit of the slope is as these two moments in time are closer and closer, and as those two moments get closer and closer, the slope comes out to be closer and closer to the slope at a single instant.

External links

• Online interactive exercises on limits
• Tutorials for the calculus phobe

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