AP Calculus AB and BC

Limit of sin x / x as x Approaches 0 Is 1

Q: Why is the limit 1 and not 0?

Because a \frac{0}{0} quotient is decided by rates, not by the fact that both parts reach 0. The graphs of y = \sin x and y = x share the tangent line y = x at the origin, so near 0 the two are nearly equal and their ratio settles at 1.

Q: Can I use L'Hopital's rule on sin x / x?

It produces the right number, \lim_{x \to 0} \frac{\cos x}{1} = 1, but it assumes the derivative of \sin x, which is derived from this very limit. Treat it as a quick check on your answer; the squeeze theorem is the honest justification.

Q: What is the limit of sin(ax)/x?

It is a for any constant a. For a \neq 0, write \frac{\sin(ax)}{x} = a \cdot \frac{\sin(ax)}{ax}, and the fraction matches the special limit since ax \to 0. For a = 0 the expression is identically 0. So \frac{\sin 2x}{x} \to 2 and \frac{\sin x}{5x} \to \frac{1}{5}.

The limit of sin x over x as x approaches 0 is 1. Direct substitution gives 0/0, so it takes the squeeze theorem or the unit-circle argument, not substitution. The result holds only in radians; in degrees the same limit is pi over 180. This limit is what makes the derivative of sin x equal cos x.

\lim_{x \to 0} \frac{\sin x}{x} = 1

Settled by the special trigonometric limit.

How the squeeze theorem settles it

Draw the unit circle and take $x$ with $0 < x < \frac{\pi}{2}$ . Three regions nest inside each other: the triangle with height $\sin x$ , the circular sector of angle $x$ , and the triangle with height $\tan x$ . Their areas are in the same order.

\frac{1}{2}\sin x < \frac{1}{2}x < \frac{1}{2}\tan x

Divide through by $\frac{1}{2}\sin x$ , which is positive on that interval, then take reciprocals (this reverses the inequalities).

1 < \frac{x}{\sin x} < \frac{1}{\cos x} \;\Longrightarrow\; \cos x < \frac{\sin x}{x} < 1

Both outer bounds go to 1 as $x \to 0$ , so the expression trapped between them has nowhere else to go. The bounds cover the left side too, because $\frac{\sin(-x)}{-x} = \frac{\sin x}{x}$ makes the function even.

\lim_{x \to 0^-} \frac{\sin x}{x} = \lim_{x \to 0^+} \frac{\sin x}{x} = 1

Why direct substitution fails

Substituting $x = 0$ gives $\frac{\sin 0}{0} = \frac{0}{0}$ , an indeterminate form. That is not a value and not a verdict of "no limit". It reports only that numerator and denominator vanish together, which leaves the answer to be decided by how fast each one does it.

Near zero the two vanish at the same rate. The Maclaurin series $\sin x = x - \frac{x^3}{6} + \cdots$ divided by $x$ gives $1 - \frac{x^2}{6} + \cdots$ , so the quotient drifts up to 1 from below.

$x$	$\frac{\sin x}{x}$
0.1	0.998334
0.01	0.999983
0.001	1.000000

Radians are not optional

The proof uses the sector area $\frac{1}{2}x$ , a formula that is only true when $x$ is a radian measure. In degrees, $\sin(x^\circ) = \sin\left(\frac{\pi x}{180}\right)$ and the limit becomes $\frac{\pi}{180} \approx 0.01745$ . A calculator left in degree mode is the usual reason a student sees 0.01745 instead of 1.

Common mistakes

Cancelling the $x$ . In $\sin x$ the $x$ is an argument, not a factor, so it cannot cancel with the denominator. $\frac{\sin x}{x}$ does not reduce to $\sin$ or to 1 by algebra.
Reading $\frac{0}{0}$ as 0, as 1, or as undefined. All three are guesses; the form carries no value of its own.
Calling L'Hopital a proof. Differentiating top and bottom gives $\frac{\cos x}{1} \to 1$ , and the number is right, but $\frac{d}{dx}[\sin x] = \cos x$ is itself proved from this limit, so the argument is circular. Use it as a check, not as a justification.
Carrying the answer to infinity. $\lim_{x \to \infty} \frac{\sin x}{x} = 0$ , because the numerator stays between $-1$ and 1 while the denominator grows. Same expression, different question.
Assuming the companion limit matches. $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ , not 1.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Why is the limit 1 and not 0?

Because a $\frac{0}{0}$ quotient is decided by rates, not by the fact that both parts reach 0. The graphs of $y = \sin x$ and $y = x$ share the tangent line $y = x$ at the origin, so near 0 the two are nearly equal and their ratio settles at 1.

Can I use L'Hopital's rule on sin x / x?

It produces the right number, $\lim_{x \to 0} \frac{\cos x}{1} = 1$ , but it assumes the derivative of $\sin x$ , which is derived from this very limit. Treat it as a quick check on your answer; the squeeze theorem is the honest justification.

What is the limit of sin(ax)/x?

It is $a$ for any constant $a$ . For $a \neq 0$ , write $\frac{\sin(ax)}{x} = a \cdot \frac{\sin(ax)}{ax}$ , and the fraction matches the special limit since $ax \to 0$ . For $a = 0$ the expression is identically 0. So $\frac{\sin 2x}{x} \to 2$ and $\frac{\sin x}{5x} \to \frac{1}{5}$ .