AP Calculus AB and BC

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

Q: Is the limit of tan x / x the same as sin x / x?

Both equal 1, for the same reason: the extra \frac{1}{\cos x} contributes a factor of 1 at x = 0. On 0 < |x| < \frac{\pi}{2} the expressions differ, with \frac{\tan x}{x} above 1 and \frac{\sin x}{x} below it. Past \frac{\pi}{2} the comparison breaks down, because \cos x changes sign and \frac{\tan x}{x} turns negative.

Q: What is the limit of tan(ax)/(bx)?

It is \frac{a}{b} for nonzero b. Rewrite as \frac{a}{b} \cdot \frac{\tan(ax)}{ax}, and the fraction tends to 1 because ax \to 0. For instance \frac{\tan 5x}{2x} \to \frac{5}{2}.

Q: What does this limit say about the graph of tan x?

That y = x is the tangent line to y = \tan x at the origin. The slope there is \sec^2 0 = 1, and the curve leaves the line only at order x^3, which is why the quotient sits so close to 1 for small x.

The limit of tan x over x as x approaches 0 is 1. Substitution gives 0/0. Writing tan x as sin x over cos x splits the expression into sin x over x times 1 over cos x, whose limits are 1 and 1. The result says that tan x and x agree to first order at the origin.

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

Settled by rewriting as sine over cosine.

Rewrite the tangent, then split

There is no separate special limit for the tangent. Every tangent limit at 0 is built from the sine one, so replace $\tan x$ with $\frac{\sin x}{\cos x}$ and regroup until $\frac{\sin x}{x}$ sits intact in one factor.

\frac{\tan x}{x} = \frac{\sin x}{x\cos x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}

The first factor is the special limit, worth 1. The second is continuous at $x = 0$ , so substitution gives $\frac{1}{\cos 0} = 1$ . Both limits exist, so the product law lets you multiply them.

\lim_{x \to 0} \frac{\tan x}{x} = 1 \cdot 1 = 1

Why direct substitution fails

At $x = 0$ the numerator is $\tan 0 = 0$ and the denominator is 0, so substitution returns the indeterminate $\frac{0}{0}$ . The expression is undefined at the point itself, which never blocks a limit: a limit only asks what happens nearby.

Nearby, the series $\tan x = x + \frac{x^3}{3} + \cdots$ makes the numerator slightly larger than the denominator, so the quotient comes down to 1 from above. That is the mirror image of $\frac{\sin x}{x}$ , which climbs to 1 from below.

$x$	$\frac{\tan x}{x}$
0.1	1.003347
0.01	1.000033
0.001	1.000000

Radian mode

The special limit behind this one is a statement about radian measure. In degrees the value is $\frac{\pi}{180} \approx 0.01745$ , which is what a calculator in the wrong mode will show for a table of values near 0.

Common mistakes

Treating $\tan x$ as a product and cancelling the $x$ . The $x$ is the angle, not a factor.
Assuming every trig expression over $x$ tends to 1. $\frac{1-\cos x}{x} \to 0$ and $\frac{\sin 2x}{x} \to 2$ ; the value 1 belongs only to forms that reduce to $\frac{\sin u}{u}$ with matching arguments.
Skipping the argument fix on relatives. $\lim_{x \to 0} \frac{\tan 3x}{x} = 3$ , because $\frac{\tan 3x}{x} = 3 \cdot \frac{\tan 3x}{3x}$ .
Forgetting that $\frac{1}{\cos x}$ needs continuity to be evaluated by substitution. It is fine at 0, but the same factor is what makes $\frac{\tan x}{x}$ blow up near $\frac{\pi}{2}$ .
Leaning on L'Hopital as the justification. $\frac{\sec^2 x}{1} \to 1$ is correct arithmetic, but the derivative of $\tan x$ is built from the quotient rule on the sine and cosine derivatives, which trace back to the special limit.

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

Is the limit of tan x / x the same as sin x / x?

Both equal 1, for the same reason: the extra $\frac{1}{\cos x}$ contributes a factor of 1 at $x = 0$ . On $0 < |x| < \frac{\pi}{2}$ the expressions differ, with $\frac{\tan x}{x}$ above 1 and $\frac{\sin x}{x}$ below it. Past $\frac{\pi}{2}$ the comparison breaks down, because $\cos x$ changes sign and $\frac{\tan x}{x}$ turns negative.

What is the limit of tan(ax)/(bx)?

It is $\frac{a}{b}$ for nonzero $b$ . Rewrite as $\frac{a}{b} \cdot \frac{\tan(ax)}{ax}$ , and the fraction tends to 1 because $ax \to 0$ . For instance $\frac{\tan 5x}{2x} \to \frac{5}{2}$ .

What does this limit say about the graph of tan x?

That $y = x$ is the tangent line to $y = \tan x$ at the origin. The slope there is $\sec^2 0 = 1$ , and the curve leaves the line only at order $x^3$ , which is why the quotient sits so close to 1 for small $x$ .