xaktly | Calculus


This section is the beginning of your studies on limits. It is related to the sections on

The notion of a limit: You've seen it before

asymptotic function

In your studies of rational functions like the one above, you've already encountered the concept of asymptotic behavior or asymptotes.

Asymptotes are imaginary lines (in this case, the horizontal line y = 0) that the graph of a function approaches – but never quite reaches – as the value of the independent variable approaches some value (± ∞ in this case).

For the function $1/(x^2+1),$ as x grows toward ± ∞ (that is, toward -∞ from the right or toward +∞ from the left), the value of the function gets closer and closer to zero, but it will never actually be zero. We say that the limit of the function as x approaches ± ∞ is zero.



An asymptote is an imaginary line (of any slope, but often vertical or horizontal) which a function approaches as the independent variable approaches some value. The function graph always approaches the line, but will never meet it.

Another example:   $f(x) = 1/x$

In this example, the simplest of all rational functions, we also have asymptotic behavior as x approaches ± ∞, but we also have a pair of infinite limits.

Notice that as the denominator approaches zero from either the right or the left, the function grows very large (1/1=1 → 1/0.1=10 → 1/0.01=100 → 1/0.001=1000 → ... ). As x→0 from the left, the function → -∞, and from the right it goes to +∞..

This idea of approaching a limit from the right and from the left will be important as we go on. Notice that this function is really undefined at x = 0, and that the limits from right and left at x = 0 are different:

From the right, the value of this function tends toward +∞ and from the left it tends toward -∞.


Why are limits important?

The concept of a limit is the fundamental basis for both flavors of calculus, differential and integral, and will be extremely important when you study how to replace very complicated functions with infinite series of simpler terms.

If you've completed your study of elementary functions, like quadratic, rational, trigonometric, and other functions, the next step in almost any other field of mathematics will require some understanding of limits.

Limit notation

There is a particular notation for writing limits that you'll need to learn and know. Consider our first function above. We said that the limit of the function   $f(x) = 1 / (x^2 + 1)$ as x approaches ± ∞; is zero.

In mathematical language, that's:

$$\lim_{x\to\infty} \; \frac{1}{x^2 + 1} = 0$$

We read this statement like this: "As the variable x gets infinitely large, the value of the function f(x) = 1 / (x2 + 1) approaches zero."

Reading and writing limit statements

We write generic limit statements like this:

$$\lim_{x\to a} \; f(x) = c$$

where f(x) is the function with independent variable x, a is the value that x approaches and c is the value that the function approaches when x is very close to a.

As the variable x in the function f(x) approaches the value a, the value of the function approaches c.

We'll make much more use of limit statements as we go on, so make sure to notice and learn them. The box below expands a bit on notation for limits approached from above (larger x) and below (smaller x).

Nothing "equals" infinity

Just one final thing about limit notation, and for me it's more of a peeve than a hard-fast rule. For example, in the second example function above, f(x) = 1/x, the limit of the function as x approaches zero from above is infinite. We wouldn't want to write

$$\lim_{x\to\infty} \; \frac{1}{x} = \infty$$

because nothing can really equal infinity. Instead, many prefer an arrow to show that the function gets larger infinitely:

$$\lim_{x\to\infty} \; \frac{1}{x} \rightarrow \infty$$

Limit notation

Limit notation: Here are three examples of limit notation and their interpretations

An example of limits – the derivative

The idea behind the derivative is to be able to find the slope of a curve , which is the slope of a curved function at any single point in its domain. That slope of a curve is defined as the slope of a line tangent to the curve at the point of interest.

Using the figure below, let's try to find the slope of the function f(x) = x2 at the point (1, 1).

As a first approximation, we can just draw some arbitrary point (x, y) and connect it to (1, 1) with a line segment (a secant, actually). We can easily calculate the slope of the resulting segment:

$$m(x) = \frac{y - 1}{x - 1} = \frac{x^2 - 1}{x - 1}, \; \color{red}{x \ne 1}$$

where we just substituted x2 for y using the function definition. Remember that slope is just "rise over run." So far, so good.

Well, that is, of course, a terrible approximation of the slope of f(x) = x2 at (1, 1). But we can do better by moving the point (x, y) closer to (1, 1), like this:

Notice that we can't keep moving our point (x, y) all the way to (1, 1) because if x = 1, our quotient "blows up" (that's math-speak for "has a zero denominator"). Well, we can work around that.

Notice that because   $x^2 - 1 = (x + 1)(x - 1),$

$$m(x) = \frac{(x + 1)(x - 1)}{x - 1} = x + 1$$

Now we've gone from a situation in which we couldn't just plug x = 1 in to one in which we can. The slope of this graph at (1, 1) is m(1) = 1 + 1 = 2. We couldn't actually find m(x) at first, but with some work and a bit of cleverness (originally discovered by Newton and Leibniz), we found its limit as x → 1.

As x goes to 1, m(x) approaches 2. The condition x ≠ 1 still holds, so while the value of m(x) can't actually get to 2, it can get arbitrarily close: 1.99999, 2.00000001, ... you get the point. Because x cannot equal 1, we can't say straight out that m(x) = 2, only that the limit of the function m(x) approaches 2 as x approaches 1, which we write in a shorthand like this:

$$\lim_{x\to 1} \; m(x) = 2$$

Such a statement, a limit statement, is read: "the limit of m(x), as x approaches a value of 1, is 2."

This procedure, in which we examine the limit of a slope function as we vary the domain variable(s), is a the heart of every discussion of derivatives we will have. The concept of limits will also return when we study integrals in calculus.

In the box below, you'll find a summary of a few different kinds of limit notation. Don't worry about memorizing them; we'll use them in the examples below and you'll see how they work.

Limit notation Meaning
$$\lim_{x\to a} \; f(x) = L$$ The limit of f(x) as x approaches x=a is L. This is the most common form of the limit statement you'll encounter.
$$\lim_{x\uparrow a} \; f(x) = L$$ The limit of f(x) as x approaches a from below (from negative to positive) is L. The limit approaching from above may be different.
$$\lim_{x\to a^-} \; f(x) = L$$ Same as above; this is now the preferred notation.
$$\lim_{x\downarrow a} \; f(x) = L$$ The limit of f(x) as x approaches x=a from above (from positive to negative) is L. The limit approaching from below may be different.
$$\lim_{x\to a^+} \; f(x) = L$$ Same as above; this is now the preferred notation.

There are easy limits and difficult limits.

We'll always be interested in the difficult limits.

There are limits that are obvious. For example, to evaluate this limit,

$$\lim_{x\to 3} \; 2x - 1$$

is very easy to evaluate. Just plug in 3 for x and the answer is 5. "Evaluating the limit" in this case is just another way to say "find the value of the function at x = 3."

But other limits are more difficult. Consider this one:

$$\lim_{x\to 3} \frac{x^2 - 9}{x - 3}$$

Now it's not so straightforward because plugging in 3 for x would yield a zero denominator. In math, we would say that the function "blows up" at x = 3. It turns out that this limit can be evaluated by noticing that x2 - 9 = (x - 3)(x + 3), in which case we know that this function has a "hole," i.e. has no value at x = 3, but is fine everywhere else.

It turns out that in calculus, every limit that we'll be interested in is a hard limit, and we'll have to develop ways to deal with them, which will include algebraic manipulations, substitutions and graphical proofs. All of that will follow.

Example: An important limit

Let's take a look at the function   $f(x) = sin(x)/x$:

This function is clearly not defined (blows up) for x = 0; the rules of algebra do not allow division by zero, so we can't evaluate the function there, even if the numerator is zero there, too. Other than that, this is a well-behaved, even interesting-looking function.

In the table below the graph, I've calculated some values of f(x) as x→0, ever more closely, from either side on the number line.

Clearly the limit — the value that the function is heading toward — is one. Moreover, it's easy to see that the limit of f(x) as x approaches zero from the left (from negative x) is the same as that from the right (positive x). Later we will use the equivalence of one-sided limits to indicate that the overall limit of a function exists.

The function f(x) = sin(x) / x will be an important one for us as we discuss the derivatives of the trigonometric functions, and there will be much more to say about it later.

Now lets look at some functions with strange behavior to develop some terminology and to help illustrate what we mean by limits.


A discontinuity in a function graph is any place where, in tracing the function with a pencil, we'd have to lift the pencil up before continuing to draw. In the sections below we'll look at the major types of discontinuities.

Jump discontinuities

Here is the graph of the function f(x) = x/|x|. It is both undefined (zero denominator) and has a discontinuity at x = 0. The function jumps (moving left to right as we cross x = 0) from -1 to 1. Think of a discontinuity as a place where you'd have to lift your pencil in order to trace the function from left to right (or right to left, for that matter).

Notice that the limit of the function as x approaches zero from the left is different from that when x approaches zero from the right (study the red equations). That means the overall limit of this function at x = 0 does not exist (DNE).

Existence of limits

In order for the limit (L) of a function f to exist at a point, c, the left and right limits must be equal:

Replaceable discontinuities

This made-up function graph shows two types of discontinuities. The first two jump discontinuities have different limits from the left and right. As we saw in the last example, the limits of the function at those points (x = -2 and x = 0) do not exist (DNE).

The third discontinuity, at x = 3, is a hole. It is called a replaceable discontinuity because the limits from right and left are equal, therefore the limit [f(x) = -1] exists. In working with such a function, we can simply pretend that the hole isn't there, being careful, of course if we need to work with the function at or in the neighborhood of the hole.

This graph is discontinuous at x = -2, 0 and 3.

An example of a replaceable discontinuity

Here is an example of a function with a replaceable discontinuity

The rational function graph shown here has two kinds of limits. The limit at x = 2 is a hole in the function. The equations below the graph show that the same binomial, (x - 2) occurs in both numerator and denominator, indicating that a hole is present instead of a vertical asymptote at x = 2. The table shows numerically how the value of the function approaches 5/3 (1.666 ...) as x→2

The limits of the function as x approaches -1 from either side are ±

While the function viewed from the right and left of x = -1 clearly tends toward different (infinite) limits, it's generally OK to say that the limit is infinite. What's not OK is to say the the limit of the function as x→ -1 is equal to infinity. Infinity is not a number. Take a look at this simple "proof":

$$\infty + 1 = \infty \; \leftarrow \; \text{Infinity doesn't get any bigger}$$

Subtract ∞ from both sides:

$$ \begin{align} (\infty + 1) - \infty &= \infty - \infty \\ \infty - \infty + 1 &= 0 \\ 1 &\ne 0 \; \; \; \leftarrow \; \text{contradiction} \end{align}$$

Infinite limits

Many functions have infinite limits – which is really no limit at all, if you think about it. Consider the function f(x) = 1/x2 in the graph below. It is undefined at x = 0, and there's no amount of algebraic manipulation that will change that.

The limits of the function as x approaches zero from the left and right are the same, f(x) → ∞. In this case we say either that the limit does not exist (after all, the function is boundless at x = 0) or that it tends toward or "goes to" infinity. Avoid writing that something "equals" infinity. Infinity can't be equaled.

Here's another example of a function with infinite limits, the natural log function (the base doesn't really matter). Because no positive base can be raised to any power to yield a negative number, the log function has only positive numbers in its domain.

But what about x = 0? When x is less than one, we're raising the base (e in this case) to a fractional power, which means taking a root. So from x = 1 downward, the result of y = ln(x) is a smaller and smaller number, but there is no value at zero. There is no power to which the base e can be raised to yield a negative number (recall that y = ln(x) means ey = x).

Here's a strange limit

The graph of sin(π/x) is very strange indeed. You should graph it yourself on a calculator or with a computer program.

The number of wiggles of the periodic function becomes infinite as the independent variable approaches zero from either side.

When you graph this function, beware that most graphing programs will give you strange artifacts when the wiggles get dense and you change the scale in order to see them. These arise from the discrete nature of the digital display: The computer is trying to graph a smooth but very detailed function on a grid of discrete pixels.

This function has no limit at x = 0.

↓   A limitation of your calculator

Your calculator—or any digital display device, for that matter—is not capable of displaying a function like y = sin(π/x) accurately. Once the wiggles in this complicated curve get too close together, your calculator has to make a decision about which of two nearby pixels to illuminate and still be true to the curve. Once the spacing is of the order of or smaller than the pixel spacing, things go awry and the curve representation loses fidelity. Below is the graph of y = sin(π/x) at different scales on a TI-84 calculator screen. Notice how they differ from the higher-resolution figure above. Even that figure has problems when the wiggle spacing gets too tight. Be careful! Your calculator has limitations.

Note: As of 2014, newer LCD screens on the TI calculators are of higher resolution, but this fundamental limitation still exists.

Properties of limits

Limits have some useful algebraic properties that you'll want to memorize.

Consider two functions, f(x) with limit

$$\lim_{x\to c} \; f(x) = L$$

and g(x) with limit

$$\lim_{x\to c} \; g(x) = K$$

where L and K are real numbers. These are the properties of limits:

Property Statement
Multiplication by a scalar $$\lim_{x\to c} \; a\cdot f(x) = a\cdot L$$
Sum or difference $$\lim_{x\to c} \; [f(x) ± g(x)] = L ± K$$
Product $$\lim_{x\to c} \; f(x) \cdot g(x) = LK$$
Quotient $$\lim_{x\to c} \; \frac{f(x)}{g(x)} = \frac{L}{K}$$
Power $$\lim_{x\to c} \; [f(x)]^n = L^n$$

$$\lim_{x\to a^+} \; g(f(x)) = g(L),$$

provided that g(L) is defined for g(x), the outer function.

Example 1

The limit of a constant function

In these examples, the functions are constant functions (pretty uninteresting as functions go, but it helps to start here).

$$\lim_{x\to 4} \; 10 = 10 \; \; \; \lim_{x\to y} \; 7 = 7$$

On the left, the function is f(x) = 10 and on the right it's f(x) = 7. The only values these functions can take on are 10 and 7, respectively, so their limits at any value of x (4 or y in the examples) are 10 and 7. Simple.

Example 2

Direct substitution

Direct substitution of the limit is always the best thing to try, unless it isn't. When it isn't is where things get tricky (and interesting), and we'll cover that in detail in the examples below. For now, take a look at these straightforward examples, limits involving   $f(x) = x^2$   and   $f(n) = 4^n:$

$$ \begin{align} \lim_{x\to 4} \; x^2 &= 4^2 = 16 \\ \\ \lim_{n\to 3} \; 4^n &= 4^3 = 64 \end{align}$$

When possible, you should use direct substitution to find limits of functions, but the most interesting and useful limit problems will never yield to direct substitution.

Example 3

Power functions

Using the power property of limits, we can solve limits like these:

$$ \begin{align} \text{If} \; \lim_{x\to 2} \, [x + 2] &= 4, \; \text{then} \\ \\ \lim_{x\to 2} \, [x + 2]^2 &= 4^2 = 16 \end{align}$$


$$ \begin{align} \text{If} \; \lim_{x\to 2} \, [x^2 + 1] &= 5, \; \text{then} \\ \\ \lim_{x\to 2} \, [x^2 + 1]^2 &= 5^2 = 25 \end{align}$$

These limits can also be determined by direct substitution. The first is [2 + 2]2 = 16, and the second is [22 + 1]2 = 52 = 25.

The power property of limits can come in very handy in finding much harder limits, as the power part can be stripped away and applied later to the limit of the inner function. For example,

$$\lim_{x\to 2} \frac{x^2 - 4x + 4}{x^2 - 4}$$

Now this limit can be evaluated by factoring the numerator and denominator (always a good thing to try):

$$ \begin{align} &= \lim_{x\to 2} \frac{(x - 2)^2}{(x - 2)(x + 2)} \\ \\ &= \lim_{x\to 2} \frac{x - 2}{x + 2} = 0 \end{align}$$

So the limit is zero. Now the power property tells us that this same function raised to any power n must also have a limit of zero.

$$\lim_{x\to 2} \left[ \frac{x^2 - 4x + 4}{x^2 - 4} \right]^n = 0$$

because 0n = 0.

Example 4

Polynomial limits: Power & sum properties

The sum property says that the limit of a sum is the sum of the limits of the individual terms. We can use that property with the power function property to find limits of polynomial functions like this one,

$$\lim_{x\to 2} [2x^2 - x + 3]$$

a sum of terms, one of which has a power. The limit is a pretty easy one, but being able to break it down like this will be useful when you determine limits of more complicated functions. the limit is

The limit of each term was solved by direct substitution of 2 for x. We always solve polynomial limits by direct substitution, because polynomial function graphs never have discontinuities. They may have a lot of wiggles, but they're always smooth curves.

Example 5

Quotient property → Rational function

In this example, the denominator of the rational function is zero at the limiting value of x (x = 3), so the limit "blows up" (or appears to).

$$\lim_{x\to 3} \; \frac{x^2 - 9}{x - 3}$$

We note the the numerator, x2 - 9, factors to (x + 3)(x - 3), so we can cancel terms like this:

$$\lim_{x\to 3} \, \frac{x^2 - 9}{x - 3} = \lim_{x\to 3} \, \frac{(x + 3)(x - 3)}{x - 3}$$

Factor the numerator to cancel the denominator — not always possible, but when it is, it can help a lot.

Then the limit is simple:

$$= \lim_{x\to 3} \, (x + 3) = 6$$

Now direct substitution works

It's found by direct substitution. Notice that what we've actually done here is removed (ignored) a replaceable discontinuity at x = 3.

The graph of our rational function has a hole at x = 3. The limit of the function at that hole is 6. There is no 3 in the domain of the function, but as the independent variable approaches 3, either from above or below, its limit is 6.

A hole is called a replaceable discontinuity because we can just substitute the point (3, 6) in for the hole and continue to work with the function.

Example 6

Dealing with radicals — one strategy

Limits like this one that involve a radical should remind you of a trick you've used from time to time:

$$\lim_{x\to 4} \frac{\sqrt{x} - 2}{x - 4}$$

This is called a "0/0 limit" because the numerator and denominator both tend toward zero as x → 4. The trick is to multiply by the ± conjugate of the numerator (or denominator if that's the problem), like this:

$$= \lim_{x\to 4} \; \frac{\sqrt{x} - 2}{x - 4} \left( \frac{\sqrt{x} + 2}{\sqrt{x} + 2} \right)$$

Cancellation often then makes these limits easier to find

$$\lim_{x\to 4} \; \frac{x - 4}{(x - 4)(\sqrt{x} + 2)} = \frac{1}{4}$$

Remember this trick – it'll come in handy!

Practice problems

Find the limit


$$\lim_{x\to -1} \; \frac{x^2 - 1}{x + 1}$$


$$\lim_{x\to -1} \; \frac{2x^2 + x - 1}{x + 1}$$


$$\lim_{x\to 0} \; \frac{(2 + x)^3 - 8}{x}$$


$$\lim_{x\to -1} \; \frac{x^3 + 1}{x + 1}$$


$$\lim_{x\to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x}$$


$$\lim_{x\to -3} \; \frac{x^2 + x - 6}{x^2 - 9}$$


$$\lim_{x\to -4} \; \frac{\sqrt{x^2 + 9} - 5}{x + 4}$$


$$\lim_{x\to 0} \; \frac{\frac{1}{3 + x} - \frac{1}{3}}{x}$$


$$\lim_{x\to 4} \frac{\sqrt{x + 5} - 3}{x - 4}$$


$$\lim_{x\to 0} \left( \frac{1}{x\sqrt{1 + x}} - \frac{1}{x} \right)$$

Video examples

Example 1

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Example 2

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Example 3

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Example 4

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Example 5

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Example 6

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On to trigonometric limits

Once you've mastered this section, the next thing to do is move on and find the limits of some important trigonometric functions. These are foundation limits for the calculus of trig. functions.

Trigonometric limits

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