Limit and Continuity of Trigonometric Functions

In this lesson we prove that sine and cosine are continuous functions, find where the other trigonometric functions are continuous, and compute two limits that are used again and again in calculus:

${\lim_{x\to0}\frac{\sin x}{x}=1}$   and   ${\lim_{x\to0}\frac{1-\cos x}{x}=0}.$

The definition of continuity and its properties are explained in the previous lesson, Properties of continuous functions.

On this page: Sine and cosine · Other trigonometric functions · The squeeze theorem · The limit of sin x / x · The limit of (1 − cos x) / x · Using the two limits · Practice

Continuity of Sine and Cosine

For an acute angle $\theta$ of a right triangle, the sine and the cosine are ratios of sides:

${\sin\theta=\frac{\text{opposite side}}{\text{hypotenuse}}},$   ${\cos\theta=\frac{\text{adjacent side}}{\text{hypotenuse}}}.$

For angles of any size we use the unit circle ${x^2+y^2=1}.$ From now on angles are measured in radians, and we write $x$ instead of $\theta.$ If the ray from the origin that makes the angle $x$ with the positive $x$-axis (counterclockwise; clockwise for negative $x$) meets the circle at the point $P,$ then $P$ has the coordinates ${(\cos x, \sin x)}.$

Here are the graphs of ${f(x)=\sin x}$ and ${g(x)=\cos x}.$ Neither of them has a break.

The limits at 0. As $x$ approaches $0,$ the point ${P(\cos x, \sin x)}$ moves along the circle towards the point $B(1, 0),$ so its coordinates approach the coordinates of $B\text{:}$

${\lim_{x\to0}\sin x=0},$   ${\lim_{x\to0}\cos x=1}.$

These are exactly the values at $0\text{:}$ ${\sin0=0}$ and ${\cos0=1}.$ Recall the definition of continuity:

By this definition, sine and cosine are continuous at $0.$

Theorem. The functions ${\sin x}$ and ${\cos x}$ are continuous at every real number.

Proof. Let $c$ be any real number and put ${h=x-c},$ so ${x=c+h}.$ Then ${x\to c}$ means the same as ${h\to0},$ and the continuity of $f$ at $c$ means that

${\lim_{h\to0}f(c+h)=f(c)}.$

The function ${\sin x}$ is defined everywhere. By the addition formula, the limit laws and the limits at $0,$

${\lim_{h\to0}\sin(c+h)}={\lim_{h\to0}(\sin c\cos h+\cos c\sin h)}$

$={\sin c\cdot\lim_{h\to0}\cos h}+{\cos c\cdot\lim_{h\to0}\sin h}={\sin c\cdot1+\cos c\cdot0}={{\sin c}.}$

In the same way, with ${\cos(c+h)}={\cos c\cos h-\sin c\sin h},$

${\lim_{h\to0}\cos(c+h)}={\cos c\cdot1-\sin c\cdot0}={{\cos c}.}$

Continuity of the Other Trigonometric Functions

The other trigonometric functions are quotients of sine and cosine:

${\tan x=\frac{\sin x}{\cos x}},$   ${\cot x=\frac{\cos x}{\sin x}},$   ${\sec x=\frac{1}{\cos x}},$   ${\csc x=\frac{1}{\sin x}}.$

A quotient of continuous functions is continuous wherever its denominator is not $0.$ The cosine is $0$ at ${x=\frac{\pi}{2}+k\pi},$ that is, at ${x=\pm\frac{\pi}{2}},$ ${\pm\frac{3\pi}{2}},$ ${\pm\frac{5\pi}{2}},$ …, and the sine is $0$ at ${x=k\pi}$ ($k$ an integer). Therefore:

Function	Continuous at every $x$ except
${\tan x},$ ${\sec x}$	${x=\frac{\pi}{2}+k\pi}$
${\cot x},$ ${\csc x}$	${x=k\pi}$

At these points the functions are not defined, and their graphs have vertical asymptotes. For example, ${\lim_{x\to(\pi/2)^-}\tan x=+\infty}$ and ${\lim_{x\to(\pi/2)^+}\tan x=-\infty}.$

At every point where a trigonometric function is continuous, its limit is found by substitution.

Example 1. Find the limits:

${\lim_{x\to\pi/4}\tan x}={\tan\frac{\pi}{4}}={1},$   ${\lim_{x\to\pi}x\cos x}={\pi\cos\pi}={-\pi},$   ${\lim_{x\to0}\cos(x^2+x)}={\cos0}={1}.$

The last limit uses the theorem on composite functions: ${\cos(x^2+x)}$ is a continuous function of the continuous function ${x^2+x}.$

Obtaining Limits by Squeezing

Now consider the limits ${\lim_{x\to0}\frac{\sin x}{x}}$ and ${\lim_{x\to0}\frac{1-\cos x}{x}}.$ Here substitution does not work: at ${x=0}$ both fractions become ${\frac{0}{0}}.$ As $x$ approaches $0,$ the numerator approaches $0,$ which pulls the fraction towards $0,$ while the denominator approaches $0,$ which makes the fraction large. Which effect wins? There is no algebraic way to cancel a common factor, as we did for rational functions. The graphs suggest the answers $1$ and $0\text{:}$

To prove such limits we use the following theorem.

Squeeze theorem. Let $f,$ $g$ and $h$ be functions such that ${g(x)\le f(x)\le h(x)}$ for all $x$ in some open interval containing the point $c,$ except possibly at $c$ itself. If ${\lim_{x\to c}g(x)}={\lim_{x\to c}h(x)}={L},$ then also ${\lim_{x\to c}f(x)=L}.$

The function $f$ is caught between $g$ and $h,$ and both of them approach the same number $L,$ so $f$ has nowhere else to go: it approaches $L$ too.

Example 2. Use the squeeze theorem to find ${\lim_{x\to0}x^2\sin^2\frac{1}{x}}.$

We cannot use the product law, because ${\lim_{x\to0}\sin^2\frac{1}{x}}$ does not exist (the function oscillates between $0$ and $1$). But for every ${x\ne0}$ we have ${-1\le\sin\frac{1}{x}\le1},$ so ${0\le\sin^2\frac{1}{x}\le1}.$ Multiplying by ${x^2\ge0}$ gives

${0}\le{x^2\sin^2\frac{1}{x}}\le{x^2}.$

Since ${\lim_{x\to0}0=\lim_{x\to0}x^2=0},$ the squeeze theorem gives ${\lim_{x\to0}x^2\sin^2\frac{1}{x}=0}.$

The same idea makes the limits of sine and cosine at $0$ precise: the segment from $P$ to the $x$-axis, of length ${|\sin x|},$ is shorter than the arc $BP,$ of length $|x|,$ so ${-|x|\le\sin x\le|x|},$ and the squeeze theorem gives ${\lim_{x\to0}\sin x=0}.$

The Limit of sin x / x

The proof uses the area of a circular sector. A sector with the angle $\theta$ (in radians) is the fraction ${\frac{\theta}{2\pi}}$ of the whole circle, whose area is ${\pi r^2},$ so its area is

${A}={\frac{\theta}{2\pi}\cdot\pi r^2}={\frac{1}{2}r^2\theta}.$

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

Proof. First let ${0\lt x\lt\frac{\pi}{2}}.$ In the unit circle take the points $O(0, 0),$ $B(1, 0),$ ${P(\cos x, \sin x)}$ and ${Q(1, \tan x)},$ the point where the ray $OP$ meets the vertical line through $B$ (this line touches the circle at $B;$ in the right triangle $OBQ$ we have ${BQ=OB\cdot\tan x=\tan x}$). The triangle $OBP$ lies inside the sector $OBP,$ which lies inside the triangle $OBQ\text{:}$

Their areas are (the base is ${OB=1};$ the heights of the triangles are $\sin x$ and ${BQ=\tan x}$):

${A_{\triangle OBP}}={\frac{1}{2}\cdot1\cdot\sin x}={\frac{1}{2}\sin x},$   ${A_{\text{sector}}}={\frac{1}{2}\cdot1^2\cdot x}={\frac{1}{2}x},$   ${A_{\triangle OBQ}}={\frac{1}{2}\cdot1\cdot\tan x}={\frac{1}{2}\tan x},$

so

${0}\lt{\frac{1}{2}\sin x}\lt{\frac{1}{2}x}\lt{\frac{1}{2}\tan x}.$

Dividing by the positive number ${\frac{1}{2}\sin x}$ gives ${1\lt\frac{x}{\sin x}\lt\frac{1}{\cos x}},$ and taking reciprocals (which reverses the inequalities) gives

${\cos x}\lt{\frac{\sin x}{x}}\lt{1}.$

If ${-\frac{\pi}{2}\lt x\lt0},$ then ${0\lt-x\lt\frac{\pi}{2}},$ so the inequality just proved holds for $-x\text{:}$ ${\cos(-x)\lt\frac{\sin(-x)}{-x}\lt1}.$ Since ${\cos(-x)=\cos x}$ and ${\frac{\sin(-x)}{-x}}={\frac{-\sin x}{-x}}={\frac{\sin x}{x}},$ this is again ${\cos x\lt\frac{\sin x}{x}\lt1}.$ So the inequality holds for every $x$ with ${0\lt|x|\lt\frac{\pi}{2}}.$ Finally, ${\lim_{x\to0}\cos x=1}$ and ${\lim_{x\to0}1=1},$ so by the squeeze theorem

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

Remember that this holds for $x$ in radians: the area formula ${\frac{1}{2}r^2\theta}$ is true only for an angle measured in radians.

The limit of (1 − cos x) / x

Theorem. ${\lim_{x\to0}\frac{1-\cos x}{x}=0}.$

Proof. Multiply the numerator and the denominator by ${1+\cos x}$ and use ${1-\cos^2x=\sin^2x}\text{:}$

${\frac{1-\cos x}{x}}={\frac{1-\cos^2x}{x(1+\cos x)}}={\frac{\sin^2x}{x(1+\cos x)}}={{\frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}}.}$

As ${x\to0},$ the first factor approaches $1$ and the second approaches ${\frac{0}{1+1}=0},$ so the limit is ${1\cdot0=0}.$

Using the two limits

Example 3. Find ${\lim_{x\to0}\frac{\tan x}{x}}.$

${\lim_{x\to0}\frac{\tan x}{x}}={\lim_{x\to0}\left(\frac{\sin x}{x}\cdot\frac{1}{\cos x}\right)}={1\cdot1}={{1}.}$

Example 4. Find ${\lim_{x\to0}\frac{\sin3x}{x}}.$

Write the fraction so that the argument of the sine is also in the denominator, and put ${t=3x}$ (${t\to0}$ when ${x\to0}$):

${\lim_{x\to0}\frac{\sin3x}{x}}={\lim_{x\to0}3\cdot\frac{\sin3x}{3x}}={3\lim_{t\to0}\frac{\sin t}{t}}={{3}.}$

Example 5. Find ${\lim_{x\to0}\frac{1-\cos x}{x^2}}.$

As in the proof above, multiply by ${1+\cos x}\text{:}$

${\frac{1-\cos x}{x^2}}={\frac{\sin^2x}{x^2(1+\cos x)}}={{\left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{1+\cos x}},}$

which approaches ${1^2\cdot\frac{1}{2}=\frac{1}{2}}.$

Practice Problems

1. Find ${\lim_{x\to\pi/6}(\sin x+\cos2x)}.$

Answer

Both functions are continuous: ${\sin\frac{\pi}{6}+\cos\frac{\pi}{3}}={\frac{1}{2}+\frac{1}{2}}={1}.$

2. Where is ${f(x)=\frac{\tan x}{x}}$ continuous?

Answer

At every $x$ except ${x=0}$ (the denominator is $0$) and ${x=\frac{\pi}{2}+k\pi}$ (the tangent is not defined). The discontinuity at $0$ is removable: the limit there is $1$ (Example 3).

3. Find ${\lim_{x\to0}\frac{\sin5x}{x}}.$

Answer

${5\lim_{x\to0}\frac{\sin5x}{5x}=5}.$

4. Find ${\lim_{x\to0}\frac{\tan2x}{x}}.$

Answer

${\frac{\tan2x}{x}}={2\cdot\frac{\sin2x}{2x}\cdot\frac{1}{\cos2x}},$ which approaches ${2\cdot1\cdot1=2}.$

5. Find ${\lim_{x\to0}\frac{\sin2x}{\sin3x}}.$

Answer

${\frac{\sin2x}{\sin3x}}={\frac{\sin2x}{2x}\cdot\frac{3x}{\sin3x}\cdot\frac{2}{3}},$ which approaches ${1\cdot1\cdot\frac{2}{3}=\frac{2}{3}}.$

6. Find ${\lim_{x\to0}\frac{1-\cos2x}{x^2}}.$

Answer

${1-\cos2x=2\sin^2x},$ so the fraction is ${2\left(\frac{\sin x}{x}\right)^2},$ which approaches $2.$

7. Use the squeeze theorem to find ${\lim_{x\to0}x\sin\frac{1}{x}}.$

Answer

${\left|x\sin\frac{1}{x}\right|\le|x|},$ so ${-|x|\le x\sin\frac{1}{x}\le|x|}.$ Both ${-|x|}$ and $|x|$ approach $0,$ so the limit is $0.$

8. Find ${\lim_{x\to+\infty}\frac{\sin x}{x}}.$

Answer

For ${x\gt0}\text{:}$ ${-\frac{1}{x}\le\frac{\sin x}{x}\le\frac{1}{x}},$ and both bounds approach $0.$ By the squeeze theorem (which also holds for ${x\to+\infty}$) the limit is $0.$

Previous lesson: Properties of continuous functions. See also Limits (Intuitive Introduction), where ${\lim_{x\to0}\frac{\sin x}{x}=1}$ was first guessed from a table.