Limit of sin(x)/x as x → 0 Equals 1: Three Proofs

Quick answer:

$$\lim_{x\to0}\frac{\sin x}{x} = 1 \qquad(x\text{ in radians})$$

You can’t just substitute \( x = 0 \): that gives \( \frac00 \), an indeterminate form. Below you’ll see what the limit means, three ways to confirm it, how it proves the derivative of sine, and how to use it on harder limits.

The intuition: for small angles, sin x ≈ x

On the unit circle, an angle of \( x \) radians cuts off an arc of length exactly \( x \). The height of the point on the circle is \( \sin x \). For a tiny angle, the arc is almost a straight vertical segment, so its length and the height are nearly the same. Their ratio gets closer and closer to 1 as the angle shrinks.

On a graph, the same fact says the line \( y = x \) is the tangent line to \( y = \sin x \) at the origin. Near 0 the two graphs are almost indistinguishable, which is exactly what the linear approximation \( \sin x \approx x \) expresses. Physicists use it constantly, for example in the small-angle pendulum formula.

1. A numerical table

\( x \)	\( \frac{\sin x}{x} \)
0.5	0.958851
0.1	0.998334
0.01	0.999983
0.001	0.9999998

The values close in on 1 from below (and the function is even, so negative \( x \) gives the same numbers). A table is good evidence, but it isn’t a proof: it can’t rule out strange behavior between the points you tried. Notice also how fast the values approach 1: each time \( x \) shrinks by a factor of 10, the gap from 1 shrinks by roughly a factor of 100. That pattern comes from the next term of the Taylor series, which makes the gap about \( \frac{x^2}{6} \).

2. The squeeze theorem proof

Take \( 0 < x < \frac\pi2 \) and draw the unit circle with the point \( P = (\cos x, \sin x) \). Three regions share the corner at the origin and the point \( A = (1, 0) \):

1. The small triangle \( OAP \) has base 1 and height \( \sin x \), so its area is \( \frac12\sin x \).
2. The sector \( OAP \) of angle \( x \) has area \( \frac12x \) (a fraction \( \frac{x}{2\pi} \) of the full area \( \pi \)).
3. The large triangle with base \( OA \) and vertical side reaching the line through \( O \) and \( P \) has height \( \tan x \), so its area is \( \frac12\tan x \).

Each region contains the one before it, so the areas are in order:

$$\frac12\sin x \;\le\; \frac12x \;\le\; \frac12\tan x$$

Divide every part by \( \frac12\sin x \), which is positive here:

$$1 \;\le\; \frac{x}{\sin x} \;\le\; \frac{1}{\cos x}$$

Take reciprocals of positive numbers, which flips the inequalities:

$$\cos x \;\le\; \frac{\sin x}{x} \;\le\; 1$$

As \( x \to 0 \), \( \cos x \to 1 \). The function is trapped between two things that both approach 1, so by the squeeze theorem the limit is 1. The same holds for \( x < 0 \) because \( \frac{\sin x}{x} \) is even: replacing \( x \) by \( -x \) changes the sign of both top and bottom. So the left-hand and right-hand limits agree.

3. L’Hôpital’s rule (and a warning)

L’Hôpital’s rule gives \( \lim\frac{\cos x}{1} = 1 \) in one line. But this proof is circular: to know that \( \frac{d}{dx}\sin x = \cos x \) you need this very limit. It’s fine for checking an answer, not for proving it.

Why this limit matters: the derivative of sine

Here is where the limit earns its fame. By the definition of the derivative and the angle-sum formula for sine:

$$\begin{aligned} \frac{d}{dx}\sin x &= \lim_{h\to0}\frac{\sin(x + h) - \sin x}{h} \\ &= \lim_{h\to0}\frac{\sin x\cos h + \cos x\sin h - \sin x}{h} \\ &= \sin x\cdot\lim_{h\to0}\frac{\cos h - 1}{h} + \cos x\cdot\lim_{h\to0}\frac{\sin h}{h} \end{aligned}$$

The first limit is 0 (shown below) and the second is our limit, 1. So \( \frac{d}{dx}\sin x = \cos x \). Every trig derivative after that, like the derivative of tan x, is built on this step.

Why radians?

In degrees, \( \frac{\sin x°}{x} \to \frac{\pi}{180} \approx 0.01745 \). That’s because \( x \) degrees is \( \frac{\pi x}{180} \) radians. The clean answer of 1 is the reason calculus uses radians, and why \( \frac{d}{dx}\sin x = \cos x \) has no extra constant.

Worked examples: related limits you can now solve

The trick in almost every problem is to make the argument of sine match the denominator.

Example 1. \( \lim_{x\to0}\frac{\sin 5x}{x} \). Multiply and divide by 5:

$$\frac{\sin 5x}{x} = 5\cdot\frac{\sin 5x}{5x} \to 5\cdot1 = 5$$

As \( x \to 0 \), \( u = 5x \to 0 \) too, so \( \frac{\sin u}{u} \to 1 \).

Example 2. \( \lim_{x\to0}\frac{\sin 3x}{\sin 2x} \). Build a \( \frac{\sin u}{u} \) on top and its reciprocal on the bottom:

$$\frac{\sin 3x}{\sin 2x} = \frac{\sin 3x}{3x}\cdot\frac{2x}{\sin 2x}\cdot\frac{3x}{2x} \to 1\cdot1\cdot\frac32 = \frac32$$

Example 3. \( \lim_{x\to0}\frac{\tan x}{x} = 1 \), since \( \frac{\tan x}{x} = \frac{\sin x}{x}\cdot\frac{1}{\cos x} \) and both factors go to 1.

Example 4. \( \lim_{x\to0}\frac{1 - \cos x}{x^2} \). Multiply by the conjugate \( 1 + \cos x \) and use \( 1 - \cos^2x = \sin^2x \):

$$\frac{1 - \cos x}{x^2} = \left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{1 + \cos x} \to 1\cdot\frac12 = \frac12$$

The same idea shows \( \lim_{x\to0}\frac{1 - \cos x}{x} = 0 \): there is one extra factor of \( x \) on top, which goes to 0.

Example 5. \( \lim_{x\to0}\frac{\sin(x^2)}{x} \). Write it as \( x\cdot\frac{\sin(x^2)}{x^2} \). The fraction goes to 1 and the \( x \) goes to 0, so the limit is 0.

Example 6. \( \lim_{x\to\pi}\frac{\sin x}{\pi - x} \). Here the input doesn’t go to 0, so substitute \( u = \pi - x \). Then \( \sin x = \sin(\pi - u) = \sin u \), and as \( x \to \pi \), \( u \to 0 \):

$$\lim_{u\to0}\frac{\sin u}{u} = 1$$

Example 7. \( \lim_{x\to0}\frac{\sin 2x}{x\cos x} \). Split the fraction into pieces whose limits you already know:

$$\frac{\sin 2x}{x\cos x} = 2\cdot\frac{\sin 2x}{2x}\cdot\frac{1}{\cos x} \to 2\cdot1\cdot1 = 2$$

This is the general strategy in one line: factor out the constant that makes the sine argument match the denominator, then let every remaining factor go to its limit separately. The product rule for limits allows this because each factor has a finite limit.

Common mistakes

• Using degrees. On a calculator set to degrees, the ratio tends to about 0.01745, not 1.
• Assuming every sine ratio is 1. \( \frac{\sin 5x}{x} \) tends to 5, not 1. The argument and the denominator must match.
• “Cancelling” the sin. \( \frac{\sin 5x}{\sin 2x} \) is not \( \frac{5x}{2x} \) by algebra. It happens to give the right limit, but only because of the \( \frac{\sin u}{u} \) argument above.
• Using it away from 0. The limit is 1 only as the argument goes to 0. As \( x \to \infty \) the answer is 0 (see below).

Don’t mix it up with x → ∞

\( \lim_{x\to\infty}\frac{\sin x}{x} = 0 \), because the top stays between \( -1 \) and 1 while the bottom grows. See limits at infinity.

Try it with the limit calculator

The limit calculator handles one-sided limits and limits at infinity too, and the L’Hôpital’s rule calculator shows the \( \frac00 \) steps if you want to check an answer that way.

Practice problems

Try these before checking the answers.

1. \( \lim_{x\to0}\frac{\sin 7x}{3x} \)
2. \( \lim_{x\to0}\frac{x}{\sin x} \)
3. \( \lim_{x\to0}\frac{\sin^2x}{x^2} \)
4. \( \lim_{x\to0}\frac{\tan 3x}{x} \)
5. \( \lim_{x\to0}\frac{1 - \cos 2x}{x^2} \)
6. \( \lim_{x\to0}\frac{\tan 3x}{\sin 5x} \)

Answers: (1) \( \frac73 \); (2) \( 1 \); (3) \( 1 \); (4) \( 3 \); (5) \( 2 \); (6) \( \frac35 \).

FAQ

Is sin(x)/x defined at 0?

No, but the limit exists. Defining the value at 0 to be 1 gives the continuous “sinc” function used in signal processing.

Does the limit exist from both sides?

Yes. The function is even, so the left-hand and right-hand limits are both 1, and the two-sided limit is 1.

What is the limit if x is in degrees?

It’s \( \frac{\pi}{180} \approx 0.01745 \). Converting degrees to radians introduces that factor.

How do you prove it without L’Hôpital’s rule?

Use the squeeze theorem with the unit-circle areas, as in proof 2. That argument doesn’t rely on knowing any derivatives.

Why isn’t the limit 0, since sin 0 = 0?

Because the bottom goes to 0 as well. A limit of the form \( \frac00 \) can come out to any value; what matters is how fast the top and bottom shrink compared with each other. Near 0, \( \sin x \) and \( x \) shrink at the same rate, so their ratio tends to 1. Compare \( x\sin\frac1x \), which really does go to 0 as \( x \to 0 \), because there the sine stays bounded while the \( x \) in front shrinks.

Where is this limit used?

It’s the key step in proving the derivative of \( \sin x \), and it appears in optics, Fourier analysis and small-angle approximations. The Taylor series \( \sin x = x - \frac{x^3}{6} + \cdots \) makes it obvious: divide by \( x \) and every term but the 1 vanishes as \( x \to 0 \).

Further reading

• Proof of Trig Limits (Paul’s Online Math Notes) — a second geometric proof of this limit and of the cosine limit.
• Computing Limits (Paul’s Online Math Notes) — the squeeze theorem and other techniques for limits that don’t allow direct substitution.