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Limit Laws and Algebraic Methods · Tutorial 213 of 1495

Tangent Quotients from Sine and Cosine Limits

Learn to evaluate tangent-based limits near zero by expressing tangent as sine over cosine and applying the established trigonometric limit laws.

Beginner 8 min read

What You'll Learn

  • Rewrite tangent quotients using \(\tan u=\sin u/\cos u\).
  • Use the sine quotient limit and the cosine limit near zero.
  • Evaluate scaled tangent limits such as tangent of a constant multiple of \(x\), divided by \(x\).
  • Compare two tangent terms with different scale factors.
  • Check when denominator and angle conditions allow the limit laws to apply.
  • Recognize and correct a missing scale factor in a tangent limit.

Why Tangent Limits Reduce to Sine and Cosine

Near zero, tangent limits can be evaluated using two trigonometric limits already established: the sine quotient limit and the cosine limit. The key is to use the definition of tangent as a quotient, rather than treating tangent as a new independent limit rule.

Throughout this tutorial, angles are measured in radians. This condition matters: the standard sine quotient limit has the stated value only when its angle is in radians. We will study limits as \(x\) approaches zero, with constant multiples of \(x\) appearing inside trigonometric functions.

Definition: Wherever \(\cos u\ne 0\), tangent is defined by \(\tan u=\frac{\sin u}{\cos u}\). Thus, to evaluate a tangent-based limit, we can rewrite each tangent using this quotient whenever the cosine denominator is nonzero.

The sine quotient limit states that \(\lim_{u\to 0}\frac{\sin u}{u}=1\). The established quadratic cosine limit states that \(\lim_{u\to 0}\frac{1-\cos u}{u^2}=\frac12\). In particular, the latter implies that \(\cos u\) tends to \(1\) as \(u\) tends to zero. These two facts will be the main ingredients.

From the Cosine Limit to a Tangent Limit

First, make explicit why the quadratic cosine limit gives the cosine limit needed here. For \(u\ne0\), write \(\cos u=1-u^2\frac{1-\cos u}{u^2}\). As \(u\to0\), the quotient tends to \(\frac12\) and \(u^2\) tends to zero, so their product tends to zero. Therefore \(\cos u\to1\). Since this limit is nonzero, \(\cos u\) is nonzero for all \(u\) sufficiently close to zero. The quotient defining tangent is consequently valid in a neighborhood of zero.

Now take \(u\ne0\) and rewrite the basic tangent quotient:

$$ \frac{\tan u}{u} = \frac{\sin u}{u\cos u} = \frac{\sin u}{u}\cdot\frac{1}{\cos u}. $$

The first factor tends to \(1\), and the second tends to \(1\), because \(\cos u\to1\). The product limit law therefore gives the tangent quotient limit.

Theorem: With angles measured in radians, \(\lim_{u\to0}\frac{\tan u}{u}=1\). This follows by writing \(\tan u/u=(\sin u/u)(1/\cos u)\), using the sine quotient limit and \(\cos u\to1\). The cosine denominator is nonzero sufficiently near zero.

This argument is short, but it explains why the result is true: tangent inherits its small-angle behavior from sine, while the cosine denominator approaches \(1\) and does not alter the limiting value.

Scaled Tangent Quotients

In applications, the angle inside tangent is often \(cx\), where \(c\) is a fixed real number. The scale factor must be kept track of. For \(c\ne0\), use the factorization

$$ \frac{\tan(cx)}{x} = \frac{\sin(cx)}{cx}\cdot\frac{c}{\cos(cx)}. $$

As \(x\to0\), the angle \(cx\) also tends to zero. The first factor tends to \(1\), and the cosine in the second factor tends to \(1\). Hence the limit is \(c\). If \(c=0\), the numerator \(\tan(0x)\) is zero for every \(x\) where the expression is defined, so the same formula for the limit gives zero.

Theorem: For every real constant \(c\), \(\lim_{x\to0}\frac{\tan(cx)}{x}=c\). For \(c\ne0\), this follows by factoring the expression into \(\frac{\sin(cx)}{cx}\cdot\frac{c}{\cos(cx)}\). For \(c=0\), the quotient is identically zero wherever defined.

A related form divides by the angle itself. If \(c\ne0\), substitute \(u=cx\) into the tangent quotient limit. As \(x\to0\), \(u\to0\), so \(\tan(cx)/(cx)\to1\). This form is useful when the scale factor is present in both the tangent argument and the denominator.

$$ \lim_{x\to0}\frac{\tan(cx)}{cx}=1 \qquad (c\ne0). $$

Worked Examples

Worked Example: A Scaled Tangent Divided by \(x\)

Evaluate \(\lim_{x\to0}\frac{\tan(5x)}{3x}\). The scale \(5\) is inside tangent, while the denominator has an additional factor \(3\). Rewrite using sine and cosine, keeping both constants visible:

$$ \frac{\tan(5x)}{3x} = \frac{\sin(5x)}{3x\cos(5x)} = \frac{5}{3}\cdot\frac{\sin(5x)}{5x}\cdot\frac{1}{\cos(5x)}. $$

As \(x\to0\), \(5x\to0\), so \(\frac{\sin(5x)}{5x}\to1\) and \(\cos(5x)\to1\). Applying the product limit law gives

$$ \lim_{x\to0}\frac{\tan(5x)}{3x} = \frac{5}{3}\cdot1\cdot1 = \frac{5}{3}. $$

Check: the general scaled form gives \(\lim_{x\to0}\tan(5x)/x=5\); dividing by the extra factor \(3\) gives \(5/3\), as calculated.

Worked Example: A Quotient of Two Tangents

Evaluate \(\lim_{x\to0}\frac{\tan(4x)}{\tan(7x)}\). For \(x\ne0\) sufficiently close to zero, rewrite each tangent and factor out the angle scales:

$$ \frac{\tan(4x)}{\tan(7x)} = \frac{\sin(4x)}{\cos(4x)} \cdot \frac{\cos(7x)}{\sin(7x)} = \frac{4}{7} \cdot \frac{\sin(4x)}{4x} \cdot \frac{7x}{\sin(7x)} \cdot \frac{\cos(7x)}{\cos(4x)}. $$

The sine quotient factors tend to \(1\): in particular, \(\sin(7x)/(7x)\to1\), so its reciprocal \(7x/\sin(7x)\to1\). Both cosine factors tend to \(1\), and their quotient tends to \(1\). Thus

$$ \lim_{x\to0}\frac{\tan(4x)}{\tan(7x)} = \frac{4}{7}\cdot1\cdot1\cdot1 = \frac{4}{7}. $$

Check: applying \(\lim_{x\to0}\tan(cx)/x=c\) separately gives limits \(4\) and \(7\) for the numerator and denominator after dividing each by \(x\). Since the denominator limit \(7\) is nonzero, their ratio is \(4/7\), agreeing with the direct calculation.

Worked Example: Tangent Compared with Sine

Evaluate \(\lim_{x\to0}\frac{\tan(2x)}{\sin(6x)}\). Rewrite tangent and introduce the angle factors needed for the sine quotient limits:

$$ \frac{\tan(2x)}{\sin(6x)} = \frac{\sin(2x)}{\cos(2x)\sin(6x)} = \frac{1}{3} \cdot \frac{\sin(2x)}{2x} \cdot \frac{6x}{\sin(6x)} \cdot \frac{1}{\cos(2x)}. $$

As \(x\to0\), the two sine quotient factors tend to \(1\), and \(\cos(2x)\to1\). Therefore,

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

Check: near zero, \(\tan(2x)\) has limiting ratio \(2\) to \(x\), while \(\sin(6x)\) has limiting ratio \(6\) to \(x\). Their ratio is \(2/6=1/3\), matching the result. The denominator is nonzero for all sufficiently small nonzero \(x\), since \(\sin(6x)/(6x)\to1\).

Why the Limit Laws Apply

The quotient and product limit laws can only be used when their conditions are met. In the expressions above, cosine denominators are safe because their values tend to \(1\), not zero. A quotient whose denominator has a nonzero limit has a limit equal to the quotient of the separate limits.

For example, in \(\tan(4x)/\tan(7x)\), the original denominator tends to zero, so it would be incorrect to use a quotient law directly on the limits of the two tangent functions. Instead, divide the numerator and denominator by \(x\) and use the limits of \(\tan(4x)/x\) and \(\tan(7x)/x\). The latter limit is \(7\ne0\), so the quotient law applies to these rescaled expressions.

1
Rewrite tangent.
Replace each \(\tan u\) with \(\sin u/\cos u\), provided the cosine is nonzero near the limiting point.
2
Expose the sine quotients.
For a nonzero constant \(c\), use \(\sin(cx)/(cx)\to1\), and retain the factor \(c\) produced by this rewrite.
3
Check denominators before taking limits.
Use a quotient law only when the denominator expression has a nonzero limit after the necessary rescaling.
4
Combine the limits and verify the scale.
Multiply or divide the resulting constant factors, then check that the answer reflects every angle and denominator factor.

A Common Scale-Factor Error

A frequent mistake is to assume that \(\tan(cx)/x\) always tends to \(1\), because \(\tan u/u\to1\). That loses the scale factor \(c\). The two quotients are not the same: \(\tan(cx)/x=c\bigl(\tan(cx)/(cx)\bigr)\) when \(c\ne0\).

Common Mistake: Dropping the Angle Scale

An incorrect calculation might claim

$$ \lim_{x\to0}\frac{\tan(3x)}{x} = \lim_{x\to0}\frac{\tan u}{u} = 1. $$

This substitutes \(u=3x\) in the numerator but fails to change the denominator: \(x=u/3\), not \(u\). The correct calculation is

$$ \frac{\tan(3x)}{x} = 3\frac{\tan(3x)}{3x} \longrightarrow 3\cdot1 = 3. $$

Check by the sine-and-cosine rewrite: \(\tan(3x)/x=3[\sin(3x)/(3x)][1/\cos(3x)]\), whose factors tend to \(3\), \(1\), and \(1\), respectively. The correct limit is \(3\), not \(1\).

Another important condition is the radian convention. The limit \(\sin u/u=1\) is a radian-based statement. Using degree measure without converting the angle would change the scale and invalidate the calculations as written.

When These Limits Are Useful

Tangent quotients frequently appear with several different constant multiples of the variable. The results here let us evaluate many of them by identifying the scale factors, rather than by estimating trigonometric values or trying to substitute zero into an indeterminate quotient. The underlying method remains the same: express tangent through sine and cosine, use the sine quotient limit, and use the fact that cosine approaches \(1\).

The method also provides a useful check on algebra. If a tangent argument is multiplied by a constant, the small-angle limit records that constant. If both numerator and denominator contain tangent terms, their scale factors determine the resulting ratio, provided the denominator scale is nonzero. Keeping those factors visible is usually the simplest way to avoid errors.

Check Your Understanding

For each question, use radians and show the scale factors before taking the limit.

  1. Evaluate \(\lim_{x\to0}\frac{\tan(6x)}{x}\).
  2. Evaluate \(\lim_{x\to0}\frac{\tan(3x)}{2x}\).
  3. Evaluate \(\lim_{x\to0}\frac{\tan(5x)}{\tan(2x)}\).
  4. In one sentence, explain why the cosine denominator in the tangent rewrite is nonzero sufficiently near zero.
  5. Identify the missing factor in the incorrect claim \(\lim_{x\to0}\tan(4x)/x=1\), and state the correct limit.

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