Deriving the Tangent Limit

Why the Tangent Limit Follows from the Sine Limit

The behavior of tangent near zero is closely connected to the behavior of sine near zero. Tangent is not a separate trigonometric mystery: wherever cosine is nonzero, it is the quotient of sine and cosine. That identity lets us derive a fundamental limit using the standard sine limit and the limit of cosine.

The angle convention matters. The limit of \(\frac{\sin x}{x}\) used here equals \(1\) when \(x\) is measured in radians. If angles are measured in degrees, the corresponding limit is different. Unless a problem explicitly says otherwise, AP Calculus trigonometric limits use radians.

Key result: When \(x\) is measured in radians, \[ \lim_{x\to 0}\frac{\tan x}{x}=1. \] The result follows by writing \(\tan x=\frac{\sin x}{\cos x}\), then using \(\lim_{x\to0}\frac{\sin x}{x}=1\) and \(\lim_{x\to0}\cos x=1\).

The Facts Needed for the Derivation

The geometric derivation of the sine limit and its scaled form give the standard result \[ \lim_{x\to0}\frac{\sin x}{x}=1 \] for radian inputs. Also, cosine is continuous at zero, so \[ \lim_{x\to0}\cos x=\cos 0=1. \] These two limits are the only trigonometric limit facts needed for the tangent limit.

For \(x\) sufficiently close to zero, \(\cos x\ne0\), because \(\cos x\to1\). Thus tangent is defined in a neighborhood of zero. In evaluating a limit at zero, it is enough that the expression be defined for \(x\) near zero with \(x\ne0\); the value of the expression at \(x=0\) does not need to exist.

Quotient limit condition: If \(f(x)\to L\) and \(g(x)\to M\ne0\) as \(x\to a\), then \[ \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{L}{M}. \] The nonzero limit of the denominator is essential. Here, the relevant denominator \(\cos x\) tends to \(1\), so the quotient law applies.

For nonzero \(x\) near zero, use the identity \(\tan x=\frac{\sin x}{\cos x}\). Then divide by \(x\), keeping the factors grouped so that the known sine limit appears.

Rewrite tangent.
For \(x\) near zero with \(x\ne0\), \(\frac{\tan x}{x}=\frac{\sin x}{x\cos x}=\frac{\sin x}{x}\cdot\frac{1}{\cos x}\).

Take the limit of each factor.
The first factor tends to \(1\). The second tends to \(1\), since \(\cos x\to1\ne0\) and the reciprocal limit law applies.

Multiply the limits.
The product limit law gives \(1\cdot1=1\).

In equations, the argument is:

\begin{aligned} \lim_{x\to0}\frac{\tan x}{x} &=\lim_{x\to0}\left(\frac{\sin x}{x}\cdot\frac{1}{\cos x}\right)\\ &=\left(\lim_{x\to0}\frac{\sin x}{x}\right) \left(\lim_{x\to0}\frac{1}{\cos x}\right)\\ &=1\cdot\frac{1}{1}\\ &=1. \end{aligned}

Each step has a condition behind it. The expression is considered for \(x\ne0\), so \(\frac{\sin x}{x}\) is defined. The cosine factor is nonzero sufficiently close to zero because its limit is \(1\). Finally, both factor limits exist, so the product limit law applies.

Conclusion: \[ \lim_{x\to0}\frac{\tan x}{x}=1 \] in radians. The essential reasoning is not that tangent and its input are equal near zero; they are not. Rather, their ratio approaches \(1\).

Worked Examples

Worked Example: The Basic Tangent Ratio

Evaluate \(\displaystyle\lim_{x\to0}\frac{\tan x}{x}\).

For \(x\ne0\) near zero, rewrite tangent as sine divided by cosine:

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

Now evaluate the limits of the two factors. The standard sine limit gives \[ \lim_{x\to0}\frac{\sin x}{x}=1. \] Continuity of cosine gives \(\lim_{x\to0}\cos x=1\), so the reciprocal limit is \(\lim_{x\to0}\frac{1}{\cos x}=1\). Therefore,

\lim_{x\to0}\frac{\tan x}{x} =1\cdot1=1.

Check: substituting the established factor limits into the rewritten expression gives \(1\cdot\frac{1}{1}=1\), consistent with the claimed limit. Direct substitution into the original quotient would instead give \(\frac{0}{0}\), which does not evaluate the limit.

Worked Example: A Scaled Tangent Ratio

Evaluate \(\displaystyle\lim_{x\to0}\frac{\tan(3x)}{6x}\).

For \(x\ne0\) near zero, use \(\tan(3x)=\frac{\sin(3x)}{\cos(3x)}\). To use the standard sine limit, arrange the denominator to contain \(3x\):

\begin{aligned} \frac{\tan(3x)}{6x} &=\frac{\sin(3x)}{6x\cos(3x)}\\ &=\frac{1}{2}\cdot\frac{\sin(3x)}{3x}\cdot\frac{1}{\cos(3x)}. \end{aligned}

As \(x\to0\), the input \(3x\to0\). The scaled sine limit gives \(\lim_{x\to0}\frac{\sin(3x)}{3x}=1\), and cosine continuity gives \(\lim_{x\to0}\cos(3x)=1\). The reciprocal factor therefore tends to \(1\). Thus,

\lim_{x\to0}\frac{\tan(3x)}{6x} =\frac{1}{2}\cdot1\cdot\frac{1}{1} =\frac{1}{2}.

Check: the decomposition has the constant factor \(\frac{1}{2}\), while each trigonometric factor tends to \(1\). Their product is \(\frac{1}{2}\), as found.

Worked Example: The Reciprocal Tangent Ratio

Evaluate \(\displaystyle\lim_{x\to0}\frac{x}{\tan x}\).

For \(x\ne0\) sufficiently close to zero, \(\tan x\ne0\): in particular, \(\sin x\ne0\) for nonzero \(x\) in a sufficiently small interval around zero, and \(\cos x\ne0\) there. Rewrite the quotient:

\begin{aligned} \frac{x}{\tan x} &=\frac{x\cos x}{\sin x}\\ &=\frac{\cos x}{\sin x/x}. \end{aligned}

The numerator tends to \(1\), and the denominator tends to \(1\) by the standard sine limit. Since the denominator limit is nonzero, the quotient limit law applies:

\lim_{x\to0}\frac{x}{\tan x} =\frac{1}{1}=1.

Check: for \(x\ne0\), \(\frac{x}{\tan x}\) is the reciprocal of \(\frac{\tan x}{x}\). The latter tends to \(1\ne0\), so the reciprocal also tends to \(\frac{1}{1}=1\), confirming the result.

Why the Quotient Structure Matters

A reliable way to evaluate this limit is to preserve the part of the expression that matches a known limit. The identity for tangent produces \(\frac{\sin x}{x}\), exactly the standard sine ratio. The remaining factor, \(\frac{1}{\cos x}\), has a simple limit because cosine tends to \(1\).

This method also makes the assumptions visible. The sine limit requires radian inputs. The reciprocal factor requires cosine not to approach zero; here it approaches \(1\). And the expression need only be defined on a punctured neighborhood of zero, not at zero itself.

The result is useful because tangent expressions often appear in limit problems even when the expression is not written as \(\frac{\tan x}{x}\). Rewriting tangent in terms of sine and cosine can reveal the standard limit structure. In a scaled expression, the same principle applies: arrange the denominator so that the angle inside the sine is also the factor in the denominator, as in \(\frac{\sin(3x)}{3x}\).

A Common Mistake: Using Degrees

The value \(1\) for \(\lim_{x\to0}\frac{\sin x}{x}\) depends on measuring \(x\) in radians. A common error is to apply that result while treating \(x\) as a degree measure. In degrees, the sine limit is not \(1\).

For example, if \(d\) is measured in degrees, then \(d\) degrees equals \(\frac{\pi d}{180}\) radians. Consequently,

\lim_{d\to0}\frac{\sin(d^\circ)}{d} = \lim_{d\to0} \left( \frac{\sin(\pi d/180)}{\pi d/180}\cdot\frac{\pi}{180} \right) = \frac{\pi}{180}.

A wrong calculation would take the degree-based ratio \(\frac{\sin(d^\circ)}{d}\), replace its limit with \(1\), and conclude that the tangent ratio in degree measure also tends to \(1\). The correction is to convert the angle: \(\tan(d^\circ)=\tan(\frac{\pi d}{180})\), so

\lim_{d\to0}\frac{\tan(d^\circ)}{d} = \lim_{d\to0} \left[ \frac{\tan(\pi d/180)}{\pi d/180}\cdot\frac{\pi}{180} \right] = 1\cdot\frac{\pi}{180} = \frac{\pi}{180}.

In AP Calculus, use radians for the standard trigonometric limits unless the problem explicitly specifies a different angle convention.

Check Your Understanding

For each question, use radians and identify the limit fact that justifies each step.

Derive \(\displaystyle\lim_{x\to0}\frac{\tan x}{x}\) by writing tangent as sine divided by cosine.
Evaluate \(\displaystyle\lim_{x\to0}\frac{\tan(5x)}{10x}\).
Evaluate \(\displaystyle\lim_{x\to0}\frac{x}{\tan x}\), and explain why the quotient limit law applies.
What limit does \(\cos x\) approach as \(x\to0\), and why is that enough to take its reciprocal near zero?
Why is it incorrect to use the radian sine limit when the input is measured in degrees?

Deriving the Tangent Limit

What You'll Learn

Why the Tangent Limit Follows from the Sine Limit

The Facts Needed for the Derivation

Deriving the Tangent Limit