Limits (Intuitive Introduction)

The limit is the central idea of calculus. Roughly speaking, the limit of a function at a point is the value that the function gets close to when $x$ gets close to that point. On this page we see where limits come from (two old geometric problems), how to guess a limit from a table or a graph, where this can go wrong, and what infinite limits and limits at infinity mean. How to compute limits exactly is explained in the next lesson, Limits and Computational Approach.

On this page: Two problems · Tangent lines · Areas · A limit from a table · One-sided limits · Numerical pitfalls · Infinite limits · Limits at infinity · Summary · Common mistakes · Practice

Two Problems That Lead to Limits

The tangent problem. Given a function $f$ and a point $P(x_0, y_0)$ on its graph, find an equation of the line tangent to the graph at $P.$

The area problem. Given a function $f$ with ${f(x)\ge0},$ find the area between the graph of $f$ and an interval $[a, b]$ on the $x$-axis.

Both problems look simple, but neither can be solved with the algebra and geometry of straight lines and polygons alone. Both are solved by the same new idea: the limit.

The tangent problem

For a circle the tangent is easy to define: a line is tangent to a circle if it meets the circle at exactly one point. Such a line is perpendicular to the radius drawn to the point of contact.

For other curves this definition does not work. In the first picture below, the horizontal line meets the parabola ${x=y^2}$ at only one point $P,$ but it is not a tangent: it does not touch the curve at $P,$ it cuts through it. The line that touches the curve at $P$ is the vertical one. In the second picture, the line touches the curve at $P$ and is tangent there, yet it meets the curve at two more points.

A definition that works for other curves too uses secant lines. Take a second point $Q$ on the curve and draw the line $PQ.$ Now let $Q$ move along the curve towards $P.$ The secant line turns, and as $Q$ gets closer and closer to $P,$ it approaches a limiting position, if there is one. The line in this limiting position is the tangent line at $P.$

Example 1. Find the tangent to the parabola ${y=x^2}$ at the point $P(1, 1).$

Take a point $Q(x, x^2)$ on the parabola, ${x\ne1}.$ The slope of the secant line $PQ$ is

${m_{PQ}}={\frac{x^2-1}{x-1}}={\frac{(x-1)(x+1)}{x-1}}={{x+1}.}$

Here are the slopes for points $Q$ closer and closer to $P,$ from both sides:

$Q$ to the right of $P$		$Q$ to the left of $P$
$x$	slope of $PQ$	$x$	slope of $PQ$
3	4	0	1
2	3	0.5	1.5
1.5	2.5	0.9	1.9
1.1	2.1	0.99	1.99
1.01	2.01	0.999	1.999
1.001	2.001	0.9999	1.9999

The closer $Q$ is to $P,$ the closer the slope is to $2.$ We say that the slope of the tangent is the limit of the slopes of the secants as $x$ approaches $1,$ and we write

$$m=\lim_{x\to1}\frac{x^2-1}{x-1}=2.$$

The tangent passes through $P(1, 1)$ and has slope $2,$ so its equation is ${y-1=2(x-1)},$ that is, ${y=2x-1}.$ Notice that the fraction ${\frac{x^2-1}{x-1}}$ has no value at ${x=1}$ (it becomes ${\frac{0}{0}}$). The limit uses only the values near $1,$ never the value at $1.$ The slope of the tangent at a point is called the derivative of the function at that point.

The area problem

The areas of some plane regions can be found by cutting them into finitely many rectangles or triangles and adding the areas of the pieces: ${A=A_1+A_2}$ in the left figure and ${A=A_1+A_2+A_3}$ in the right one.

For a region with a curved side, such as the region under the graph of $f$ between ${x=a}$ and ${x=b},$ this is impossible: no finite number of rectangles or triangles fits it exactly.

Instead, we approximate the area. We divide $[a, b]$ into $n$ parts of equal width and put a rectangle on each part, as high as possible but still under the curve (an inscribed rectangle). The sum of the areas of these rectangles is a little smaller than the area of the region. As $n$ grows, the rectangles get narrower, the gaps between them and the curve get smaller, and the sums approach the exact area as a limiting value.

Example 2. Find the area under ${y=x^2}$ between ${x=0}$ and ${x=1}.$

Divide $[0, 1]$ into $n$ parts of width ${\frac{1}{n}}.$ The inscribed rectangles have the heights ${(k/n)^2}$ for ${k=0},$ $1,$ …, ${n-1}.$ Rectangles that cover the region have the heights ${(k/n)^2}$ for ${k=1},$ $2,$ …, $n;$ their total area is a little larger than the area of the region. Using the formula

$$1^2+2^2+\ldots+m^2=\frac{m(m+1)(2m+1)}{6},$$

we find the sum $s_n$ of the inscribed rectangles and the sum $S_n$ of the covering ones:

${s_n=\frac{(n-1)(2n-1)}{6n^2}}$   and   ${S_n=\frac{(n+1)(2n+1)}{6n^2}}.$

$n$	$s_n$ (below the curve)	$S_n$ (covering the region)
4	0.21875	0.46875
10	0.285	0.385
100	0.32835	0.33835
1000	0.3328335	0.3338335

For every $n$ the area lies between $s_n$ and $S_n,$ and both sums approach ${\frac{1}{3}}.$ So the area is exactly ${\frac{1}{3}}.$ A number obtained as such a limit of sums is called a (definite) integral; here it is the integral of $x^2$ from $0$ to $1.$

Guessing a Limit from a Table

Let us take

$$f(x)=\frac{\sin x}{x},$$

where $x$ is in radians. This function is not defined at ${x=0},$ because there the fraction becomes ${\frac{0}{0}}.$ What happens to $f(x)$ when $x$ is close to $0?$ Using a calculator set to radian mode, we get:

$x$	${\frac{\sin x}{x}}$	$x$	${\frac{\sin x}{x}}$
1.0	0.84147	−1.0	0.84147
0.8	0.89670	−0.8	0.89670
0.6	0.94107	−0.6	0.94107
0.4	0.97355	−0.4	0.97355
0.2	0.99335	−0.2	0.99335
0.01	0.99998	−0.01	0.99998

The values for $x$ and $-x$ are equal, because ${\frac{\sin(-x)}{-x}}={\frac{-\sin x}{-x}}={\frac{\sin x}{x}}.$ As $x$ approaches $0$ from the left or from the right, $f(x)$ approaches $1.$

We say that the limit of $f(x)$ as $x$ approaches $0$ is $1,$ and we write

$$\lim_{x\to0}\frac{\sin x}{x}=1.$$

The table only suggests this result; it is proved in calculus, and it is one of the most important limits. Note that $f(0)$ is not defined: the graph has a hole at $(0, 1).$ The limit describes the values of $f$ near $0,$ not at $0.$

Limit (informal definition). We write ${\lim_{x\to a}f(x)=L}$ and say "the limit of $f(x)$ as $x$ approaches $a$ is $L$" if the values $f(x)$ can be made as close to $L$ as we like by taking $x$ close enough to $a,$ but not equal to $a.$

The function does not even have to be defined at $a,$ only at the points near $a.$ And if it is defined there, the value $f(a)$ plays no role: changing $f(a)$ does not change the limit.

One-Sided and Two-Sided Limits

Sometimes it matters from which side $x$ approaches $a.$ If $x$ approaches $a$ through values larger than $a$ (from the right), we write ${x\to a^+}.$ If $x$ approaches $a$ through values smaller than $a$ (from the left), we write ${x\to a^-}.$

Notation	How to read the notation
${\lim_{x\to a^+}f(x)=L_1}$	The limit of $f(x)$ as $x$ approaches $a$ from the right is $L_1$ (the right-hand limit)
${\lim_{x\to a^-}f(x)=L_2}$	The limit of $f(x)$ as $x$ approaches $a$ from the left is $L_2$ (the left-hand limit)
${\lim_{x\to a}f(x)=L}$	The limit of $f(x)$ as $x$ approaches $a$ is $L$ (the two-sided limit)

For the function above, the table suggests ${\lim_{x\to0^+}\frac{\sin x}{x}=1}$ (right-hand limit) and ${\lim_{x\to0^-}\frac{\sin x}{x}=1}$ (left-hand limit).

The two-sided limit ${\lim_{x\to a}f(x)}$ exists and equals $L$ if and only if both one-sided limits exist and are equal to $L.$

Example 3. Let ${f(x)=\frac{|x|}{x}}.$ For ${x\gt0}$ we have ${|x|=x},$ so ${f(x)=1}.$ For ${x\lt0}$ we have ${|x|=-x},$ so ${f(x)=-1}.$ Therefore

${\lim_{x\to0^+}\frac{|x|}{x}=1},$   ${\lim_{x\to0^-}\frac{|x|}{x}=-1}.$

The one-sided limits are different, so the limit ${\lim_{x\to0}\frac{|x|}{x}}$ does not exist. The graph "jumps" at $0.$

Numerical Pitfalls

A table of values can only suggest a limit, and it can suggest a wrong one. Consider

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

For ${x=1},$ $0.1,$ $0.01,$ … the number ${\frac{\pi}{x}}$ is $\pi,$ $10\pi,$ $100\pi,$ …, an integer multiple of $\pi,$ and the sine of such a number is $0\text{:}$

$x$	$f(x)$	$x$	$f(x)$
1	0	−1	0
0.1	0	−0.1	0
0.01	0	−0.01	0
0.001	0	−0.001	0
0.0001	0	−0.0001	0

The table suggests that ${\lim_{x\to0}\sin\frac{\pi}{x}=0}.$ This is wrong. For ${x=\frac{2}{5}},$ ${\frac{2}{9}},$ ${\frac{2}{13}},$ …, that is, for ${x=\frac{2}{4k+1}},$ we get ${\frac{\pi}{x}}={\frac{\pi}{2}+2k\pi},$ and the sine of this number is $1.$ These values of $x$ (${k=1},$ $2,$ $3,$ …) also get as close to $0$ as we like. The same happens on the left of $0\text{:}$ for ${x=-\frac{2}{4k+3}}$ (${-\frac{2}{3}},$ ${-\frac{2}{7}},$ …) we get ${\frac{\pi}{x}}={-\frac{3\pi}{2}-2k\pi},$ and the sine is again $1.$ So, near $0,$ the function takes the value $0$ and the value $1$ (and also $-1$) again and again: its graph oscillates between $-1$ and $1$ infinitely many times. The values do not approach any single number, so ${\lim_{x\to0^+}\sin\frac{\pi}{x}},$ ${\lim_{x\to0^-}\sin\frac{\pi}{x}}$ and ${\lim_{x\to0}\sin\frac{\pi}{x}}$ do not exist.

At the red points (${x=\pm1},$ ${\pm\frac{1}{2}},$ ${\pm\frac{1}{3}},$ ${\pm\frac{1}{10}}$) ${f(x)=0},$ as in the table; at the green points ${f(x)=1}.$ Near $0$ the oscillations are too dense to draw.

A table, a calculator or a computer plot can mislead us. A guess from a table should be checked with algebra or with a careful look at the function.

Infinite Limits

Consider ${f(x)=\frac{1}{(x-2)^2}}.$ When $x$ is close to $2,$ from either side, ${(x-2)^2}$ is a small positive number, so $f(x)$ is a large positive number. For example, ${f(2.01)}={f(1.99)}={10\,000}$ and ${f(2.001)}={1\,000\,000}.$ The values grow without bound, and we write

${\lim_{x\to2^-}f(x)}={\lim_{x\to2^+}f(x)}={\lim_{x\to2}f(x)}={{+\infty}.}$

Similarly, for ${g(x)=-\frac{1}{(x-2)^2}}$ the values decrease without bound on both sides of $2,$ and ${\lim_{x\to2}g(x)=-\infty}.$

The function ${h(x)=\frac{1}{x-2}}$ behaves differently on the two sides. For $x$ a little larger than $2,$ ${x-2}$ is a small positive number and $h(x)$ is a large positive number. For $x$ a little smaller than $2,$ ${x-2}$ is a small negative number and $h(x)$ is a large negative number:

${\lim_{x\to2^-}h(x)=-\infty},$   ${\lim_{x\to2^+}h(x)=+\infty}.$

The symbols $+\infty$ and $-\infty$ are not numbers. Writing ${\lim_{x\to2}f(x)=+\infty}$ does not mean that the limit exists; it tells us how it fails to exist: the values grow beyond every bound. For $h$ the one-sided limits are different, so ${\lim_{x\to2}h(x)}$ is neither $+\infty$ nor $-\infty;$ we only say that it does not exist.

In all three pictures the dashed line ${x=2}$ is a vertical asymptote. In general, the line ${x=a}$ is a vertical asymptote of the graph of $f$ if at least one of the one-sided limits of $f$ at $a$ is $+\infty$ or $-\infty.$

Limits at Infinity

We can also ask what happens to $f(x)$ when $x$ grows without bound (${x\to+\infty}$) or decreases without bound (${x\to-\infty}$). We write ${\lim_{x\to+\infty}f(x)=L}$ if $f(x)$ can be made as close to $L$ as we like by taking $x$ large enough, and similarly for ${x\to-\infty}.$ Then the line ${y=L}$ is a horizontal asymptote of the graph.

Example 4. For the function in the picture below

${\lim_{x\to-\infty}f(x)=-1},$   ${\lim_{x\to+\infty}f(x)=4}.$

Both limits exist. They are different, but that is no problem: they describe two different ends of the graph. The lines ${y=-1}$ and ${y=4}$ are horizontal asymptotes.

Example 5. For the function below, as ${x\to-\infty},$ ${f(x)\to+\infty},$ that is, ${\lim_{x\to-\infty}f(x)=+\infty}.$ As ${x\to+\infty},$ the graph oscillates around the line ${y=-2},$ but the swings get smaller and smaller, so $f(x)$ approaches $-2\text{:}$

${\lim_{x\to-\infty}f(x)=+\infty},$   ${\lim_{x\to+\infty}f(x)=-2}.$

The line ${y=-2}$ is a horizontal asymptote, although the graph crosses it again and again.

Example 6. For the function below, as ${x\to-\infty},$ ${f(x)\to-\infty}.$ As ${x\to+\infty},$ $f(x)$ keeps oscillating between $-3$ and $3$ and does not approach any number. So ${\lim_{x\to-\infty}f(x)=-\infty},$ and ${\lim_{x\to+\infty}f(x)}$ does not exist (it is not $+\infty$ or $-\infty$ either). This graph has no horizontal asymptote.

Summary of the Notation

Notation	Meaning
${\lim_{x\to a}f(x)=L}$	$f(x)$ is as close to $L$ as we like for all ${x\ne a}$ close enough to $a$
${\lim_{x\to a^+}f(x)=L}$	the same, but only for ${x\gt a}$
${\lim_{x\to a^-}f(x)=L}$	the same, but only for ${x\lt a}$
${\lim_{x\to a}f(x)=+\infty}$	$f(x)$ is larger than any given number for all ${x\ne a}$ close enough to $a$
${\lim_{x\to a}f(x)=-\infty}$	$f(x)$ is smaller than any given number for all ${x\ne a}$ close enough to $a$
${\lim_{x\to+\infty}f(x)=L}$	$f(x)$ is as close to $L$ as we like for all large enough $x$
${\lim_{x\to-\infty}f(x)=L}$	$f(x)$ is as close to $L$ as we like for all negative $x$ with large enough $|x|$

Common Mistakes

Mistake	Wrong	Right
The limit taken as the value at the point	${\frac{\sin x}{x}}$ is not defined at $0,$ so ${\lim_{x\to0}\frac{\sin x}{x}}$ does not exist	${\lim_{x\to0}\frac{\sin x}{x}=1}\text{:}$ the limit uses the values near $0,$ not at $0$
Trusting a table	${\sin\frac{\pi}{x}=0}$ for ${x=0.1},$ $0.01,$ $0.001,$ so the limit at $0$ is $0$	The limit does not exist: for ${x=\frac{2}{4k+1}}$ the value is $1$
Looking at one side only	${\lim_{x\to0^+}\frac{|x|}{x}=1},$ so ${\lim_{x\to0}\frac{|x|}{x}=1}$	The left-hand limit is $-1,$ so the two-sided limit does not exist
Writing $\infty$ for different one-sided limits	${\lim_{x\to2}\frac{1}{x-2}=\infty}$	The one-sided limits are $-\infty$ and $+\infty;$ the limit does not exist
"A graph never crosses its asymptote"	The graph of $f$ crosses ${y=-2},$ so ${\lim_{x\to+\infty}f(x)\ne-2}$	A graph may cross its horizontal asymptote, even infinitely many times (Example 5)
Degrees instead of radians	In degree mode the calculator shows ${\frac{\sin x}{x}\approx0.01745}$ near $0,$ so the limit is not $1$	${\lim_{x\to0}\frac{\sin x}{x}=1}$ holds for $x$ in radians; in degrees the limit is ${\frac{\pi}{180}\approx0.01745}$

Practice Problems

1. Using a calculator in radian mode, guess ${\lim_{x\to0}\frac{1-\cos x}{x^2}}$ from the values at ${x=0.5},$ $0.1$ and $0.01.$

Answer

The values are about $0.489670,$ $0.499583$ and $0.499996$ (the same for $-x$). The limit seems to be ${\frac{1}{2}},$ and this is indeed its value.

2. Using secant lines, find the slope of the tangent to ${y=x^2}$ at $P(3, 9),$ and an equation of the tangent.

Answer

For ${x\ne3}$ the slope of $PQ$ is ${\frac{x^2-9}{x-3}}={x+3},$ which approaches $6$ as ${x\to3}.$ The tangent is ${y-9=6(x-3)},$ that is, ${y=6x-9}.$

3. Approximate the area under ${y=x}$ between ${x=0}$ and ${x=1}$ with $n$ inscribed rectangles. Their total area is ${\frac{n-1}{2n}}.$ Compute it for ${n=10},$ $100,$ $1000.$ What is the limit?

Answer

$0.45,$ $0.495,$ $0.4995.$ The limit is ${\frac{1}{2}},$ the area of the right triangle with legs $1$ and $1.$

4. Let ${f(x)=\frac{x^2-x}{|x-1|}}.$ Find ${\lim_{x\to1^+}f(x)}$ and ${\lim_{x\to1^-}f(x)}.$ Does ${\lim_{x\to1}f(x)}$ exist?

Answer

${x^2-x=x(x-1)}.$ For ${x\gt1}\text{:}$ ${f(x)=x},$ which approaches $1.$ For ${x\lt1}\text{:}$ ${f(x)=-x},$ which approaches $-1.$ The one-sided limits are $1$ and $-1,$ so the limit does not exist.

5. The function $g$ is defined by ${g(x)=x+1}$ for ${x\lt2},$ ${g(x)=5-x}$ for ${x\gt2},$ and ${g(2)=7}.$ Find ${\lim_{x\to2^-}g(x)},$ ${\lim_{x\to2^+}g(x)},$ ${\lim_{x\to2}g(x)}$ and $g(2).$

Answer

$3,$ $3,$ $3$ and $7.$ The limit exists and equals $3,$ although ${g(2)=7}\text{:}$ the value at the point does not matter.

6. Find ${\lim_{x\to3^-}\frac{1}{x-3}},$ ${\lim_{x\to3^+}\frac{1}{x-3}}$ and ${\lim_{x\to3}\frac{1}{(x-3)^2}}.$

Answer

$-\infty,$ $+\infty$ and $+\infty.$ The line ${x=3}$ is a vertical asymptote of both graphs.

7. Find ${\lim_{x\to+\infty}\frac{1}{x}}$ and ${\lim_{x\to-\infty}\frac{1}{x}}.$ Which line is a horizontal asymptote of ${y=\frac{1}{x}}?$

Answer

Both limits are $0$ (for example, ${\frac{1}{1000}=0.001}$ and ${\frac{1}{-1000}=-0.001}$). The horizontal asymptote is the $x$-axis, ${y=0}.$

8. Guess ${\lim_{x\to+\infty}\frac{2x+1}{x}}$ from the values at ${x=10},$ $100,$ $1000,$ then explain the result.

Answer

$2.1,$ $2.01,$ $2.001\text{:}$ the limit is $2,$ because ${\frac{2x+1}{x}}={2+\frac{1}{x}}$ and ${\frac{1}{x}}$ approaches $0.$

9. Does ${\lim_{x\to0}\cos\frac{\pi}{x}}$ exist?

Answer

No. For ${x=\frac{1}{2k}}$ the value is ${\cos2k\pi=1},$ and for ${x=\frac{1}{2k+1}}$ it is ${\cos(2k+1)\pi=-1}.$ Both kinds of $x$ get as close to $0$ as we like, so the values do not approach a single number.

Next lesson: Limits and Computational Approach, where limits are computed exactly with the limit laws. Limits of sequences are explained in Limits of sequences.