Why trigonometric zero-over-zero limits need care
Trigonometric expressions can change rapidly near a point, but their derivatives are familiar: the derivative of \(\sin x\) is \(\cos x\), and the derivative of \(\cos x\) is \(-\sin x\). When a quotient of trigonometric functions approaches zero over zero, L’Hôpital’s Rule can turn the original limit into a limit involving these derivatives.
The key is to use the rule only when its hypotheses are satisfied, and to differentiate each expression correctly. In particular, a scaled argument such as \(5x\) contributes a factor of \(5\) through the chain rule. Angles in the derivative formulas here are measured in radians.
The rule and its conditions
For limits at a finite point, the version of L’Hôpital’s Rule needed here can be stated as follows. Suppose \(f\) and \(g\) are continuous at \(a\), differentiable on a punctured interval around \(a\), and \(f(a)=g(a)=0\). Suppose also that \(g'(x)\ne 0\) for \(x\ne a\) sufficiently close to \(a\). If the limit of \(f'(x)/g'(x)\) exists as \(x\) approaches \(a\), then the limit of \(f(x)/g(x)\) exists and equals it. The rule applies from either side when the corresponding hypotheses and derivative-quotient limit hold on that side.
For trigonometric limits, \(f\) and \(g\) are often compositions such as \(\sin(3x)\) or \(\tan(2x)\). The chain rule is part of the differentiation: \(\frac{d}{dx}\sin(kx)=k\cos(kx)\) and \(\frac{d}{dx}\cos(kx)=-k\sin(kx)\), for a constant \(k\). For tangent, \(\frac{d}{dx}\tan(kx)=k\sec^2(kx)\) wherever the expression is defined.
The rule concerns the limit of \(f/g\); it does not say that the derivative of a quotient is \(f'/g'\). Nor does a \(0/0\) substitution alone guarantee that the rule applies. The differentiability, denominator-derivative, and derivative-quotient conditions matter.
Why the derivative quotient gives the limit
The connection between the two quotients comes from the Cauchy Mean Value Theorem. For a point \(x\ne a\) close to \(a\), apply that theorem to \(f\) and \(g\) on the interval with endpoints \(a\) and \(x\). Since \(f(a)=g(a)=0\), it gives a point \(c\) between \(a\) and \(x\) such that
Here \(g(x)\ne 0\) is required for the quotient on the left to be defined, and \(g'(c)\ne 0\) is required for the quotient on the right. As \(x\) approaches \(a\), the point \(c\), lying between \(x\) and \(a\), also approaches \(a\). Thus, if \(f'(t)/g'(t)\) tends to a limit \(L\) as \(t\) approaches \(a\), the displayed equality shows why \(f(x)/g(x)\) tends to the same \(L\). This is the mechanism behind the rule.
One application with scaled arguments
Worked Example: A sine quotient with different scales
Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(5x)}{\tan(2x)}\).
At \(x=0\), the numerator is \(\sin 0=0\), and the denominator is \(\tan 0=0\). Both functions are differentiable near \(0\), and the derivative of the denominator, \(2\sec^2(2x)\), is nonzero there. L’Hôpital’s Rule gives
Now substitute \(x=0\) into the derivative quotient: \(\cos(0)=1\) and \(\sec^2(0)=1\). Therefore,
So the original limit is \(\frac{5}{2}\). As a check, the differentiated numerator is \(5\cos(5x)\), and the differentiated denominator is \(2\sec^2(2x)\); both chain-rule factors are present, and their values at \(0\) give exactly \(\frac{5}{2}\).
When the derivative quotient is still zero over zero
Sometimes one application does not produce a limit that can be evaluated by direct substitution. If the new quotient is again \(0/0\), another application may be appropriate, provided the hypotheses hold again. Each differentiation must be performed on the current numerator and current denominator separately.
Worked Example: A cosine difference over a squared sine
Evaluate \(\displaystyle \lim_{x\to 0}\frac{1-\cos(3x)}{\sin^2(2x)}\).
Both numerator and denominator tend to \(0\). The first derivatives are \(3\sin(3x)\) and \(4\sin(2x)\cos(2x)\), respectively. The denominator derivative is nonzero for sufficiently small nonzero \(x\), so the first application gives
At \(x=0\), the new numerator and denominator are again both \(0\). Differentiate each once more. The numerator derivative is \(9\cos(3x)\). For the denominator, the product rule gives
The second derivative quotient has a denominator that tends to \(8\), so it is nonzero near \(0\). A second application yields
Thus the original limit is \(\frac{9}{8}\). The derivative calculations can be checked directly: differentiating \(1-\cos(3x)\) gives \(3\sin(3x)\), and differentiating that gives \(9\cos(3x)\). The first denominator derivative is \(4\sin(2x)\cos(2x)\), whose derivative is the displayed second derivative. Substitution into the final quotient confirms the stated value.
Repeated applications and higher-order cancellation
A quotient can remain indeterminate after more than one application. In that situation, do not assume that the same form or the same conditions will persist automatically. After each application, check the new numerator and denominator at the point, and check that the derivative of the denominator is nonzero nearby before applying the rule again.
Worked Example: Tangent minus sine
Evaluate \(\displaystyle \lim_{x\to 0}\frac{\tan x-\sin x}{x^3}\).
The numerator and denominator both tend to \(0\). After one application, the quotient is
This is still \(0/0\), since \(\sec^2(0)-\cos(0)=1-1=0\). The denominator derivative is \(6x\), which is nonzero for nonzero \(x\) near \(0\). Differentiating again gives
The new numerator and denominator still tend to \(0\). The denominator derivative is \(6\), so a third application is valid. Differentiating the numerator carefully gives
Therefore,
The original limit is \(\frac{1}{2}\). To check the final differentiation, the product rule gives \(\frac{d}{dx}(\sec^2 x\tan x)=2\sec^2 x\tan^2 x+\sec^4 x\); multiplying by \(2\) and adding the derivative \(\cos x\) of \(\sin x\) gives the displayed numerator. The denominator has become the constant \(6\), and direct substitution gives the claimed value.
A common chain-rule mistake
When the argument of a trigonometric function is scaled, omitting the inner derivative changes the answer. For example, consider \(\displaystyle \lim_{x\to 0}\frac{\sin(4x)}{\sin(7x)}\). Both functions tend to \(0\), so L’Hôpital’s Rule applies. A common incorrect calculation is
The error is that both chain-rule factors have been omitted. The correct derivatives are \(4\cos(4x)\) and \(7\cos(7x)\), so the limit is
As a check, substituting \(x=0\) into the correctly differentiated quotient gives \(4\cos(0)/(7\cos(0))=4/7\). Keeping the scale factors is essential: they need not cancel.
Practical checks before each application
For a trigonometric quotient, use this sequence of checks:
Confirm that the numerator and denominator both tend to zero. If they do not, this is not the \(0/0\) form addressed here.
Verify differentiability near the point and that the denominator derivative does not vanish on the punctured interval being used.
Use the chain rule for scaled arguments and the product rule when needed. Do not differentiate the quotient as a single expression.
Substitute into the new quotient. Apply the rule again only if it is still an indeterminate \(0/0\) form and the conditions hold.
Why these limits matter
Trigonometric \(0/0\) limits show how local rates of change determine the relative behavior of two functions that both vanish at a point. A scaled sine or tangent can approach zero at a rate affected by its inner factor, while repeated differentiation can reveal the first nonzero contribution when earlier derivatives also vanish. The calculations are useful only when their logic is controlled: the form must be indeterminate, each derivative must be correct, and each use of the rule must meet its conditions.
Check Your Understanding
For each question, state any indeterminate form you use and show the derivative factors from scaled arguments.
- Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(3x)}{\sin(8x)}\).
- For \(\displaystyle \lim_{x\to 0}\frac{1-\cos(2x)}{x^2}\), how many applications of L’Hôpital’s Rule are needed before substitution gives a value?
- Differentiate \(\sin(6x)\) and \(\tan(3x)\), showing the chain-rule factors.
- After one application to a \(0/0\) quotient, what must be checked before applying L’Hôpital’s Rule again?
- Explain why obtaining \(0/0\) by substitution does not, by itself, justify using the rule.
