Why Inverse-Trigonometric Quotients Need Care
Quotients involving inverse trigonometric functions often have a simple limit, but the path to that limit depends on two checks: the expression must be defined on the side from which \(x\) approaches, and the quotient must actually have an indeterminate form before L’Hôpital’s rule is used.
For example, \(\arcsin x\) is defined for \(-1\le x\le 1\), while its derivative is defined only for \(-1<x<1\). Thus, a limit as \(x\to1\) involving \(\arcsin x\) is naturally a left-hand limit. At the endpoint itself, the derivative formula cannot be evaluated, but it may still be applied on the open interval immediately to the left.
This tutorial focuses on finite points: the familiar quotient \(\arcsin x/x\) near zero, related quotients at interior points, and limits approaching the endpoints of an inverse-trigonometric function’s domain.
Derivative Formulas and Their Domains
For \(-1<x<1\), the inverse sine and inverse cosine derivatives are
For every real \(x\), the inverse tangent derivative is
The square root in the first two formulas is positive on \((-1,1)\). In particular, those formulas do not give finite derivative values at \(x=1\) or \(x=-1\).
When L’Hôpital’s Rule Applies
For the quotient limits here, the relevant form of L’Hôpital’s rule is the following. Suppose \(f\) and \(g\) are differentiable on an open interval on one side of a finite point \(a\), with \(g'(x)\ne0\) there. If \(f(x)\to0\) and \(g(x)\to0\) as \(x\to a\) from that side, and if \(f'(x)/g'(x)\) tends to a limit \(L\) (finite or infinite), then \(f(x)/g(x)\to L\) from the same side. The functions need not be differentiable at \(a\) itself.
The one-sided wording matters at a domain endpoint. Before differentiating, verify that the original quotient is defined for \(x\) on the approach side and that its numerator and denominator both tend to zero. A quotient with a nonzero numerator limit is not a zero-over-zero form, even if the denominator tends to zero.
Worked Examples Near Interior Points
Worked Example: The Basic Arcsine Quotient
Evaluate \(\displaystyle\lim_{x\to0}\frac{\arcsin x}{x}\).
For \(x\) near zero, \(\arcsin x\) is defined. As \(x\to0\), continuity gives \(\arcsin x\to0\), and the denominator also tends to zero. This is a zero-over-zero form. The derivative ratio is defined for \(-1<x<1\), \(x\ne0\), so L’Hôpital’s rule applies:
As a check, the derivative of \(\arcsin x\) at zero is \(1/\sqrt{1-0^2}=1\); the quotient is precisely the difference quotient for that derivative because \(\arcsin 0=0\). The answer agrees.
Worked Example: An Arctangent Quotient with a Scale Factor
Evaluate \(\displaystyle\lim_{x\to0}\frac{\arctan(2x)}{x}\).
The numerator tends to \(\arctan 0=0\), and the denominator tends to zero. The functions are defined near zero, so we have a valid zero-over-zero form. Differentiating the numerator by the chain rule gives \(2/(1+4x^2)\), while the derivative of the denominator is \(1\). Hence
To check the scale factor, write the quotient for \(x\ne0\) as \(2\bigl(\arctan(2x)/(2x)\bigr)\). The inner quotient tends to \(1\), as is also seen by differentiating \(\arctan u\) at \(u=0\). Thus the result \(2\) is consistent.
Worked Example: A Centered Arccosine Quotient
Evaluate \(\displaystyle\lim_{x\to0}\frac{\arccos x-\pi/2}{x}\).
Since \(\arccos 0=\pi/2\), both numerator and denominator tend to zero. The quotient is defined on both sides of zero. Applying L’Hôpital’s rule gives
As a check on the sign, \(\arccos x\) decreases as \(x\) increases. Near zero, the numerator changes in the opposite direction to \(x\), which is consistent with a negative quotient limit. The derivative at zero is exactly \(-1\), matching the computed value.
One-Sided Limits at a Domain Endpoint
The domain of \(\arcsin x\) ends at \(x=1\), so an approach to that endpoint from within the domain must use \(x\to1^-\). Expressions involving \(\sqrt{1-x^2}\) in the derivative ratio can become unbounded there. That behavior may produce an infinite limit, or it may be balanced by a denominator whose derivative also becomes unbounded. The quotient must be evaluated rather than guessed from a single derivative.
Worked Example: A Quotient That Diverges at \(x=1\)
Evaluate \(\displaystyle\lim_{x\to1^-}\frac{\pi/2-\arcsin x}{1-x}\).
For \(x<1\) near \(1\), \(\arcsin x\) is defined. As \(x\to1^-\), \(\arcsin x\to\pi/2\), so both numerator and denominator tend to zero. The derivatives exist on the open interval near \(1\), even though the arcsine derivative is not finite at \(1\). L’Hôpital’s rule yields
Indeed, \(1-x^2=(1-x)(1+x)\to0^+\), so its positive square root tends to zero and its reciprocal tends to \(+\infty\). The original numerator and denominator are both positive for \(x<1\) near \(1\), which checks the positive sign of the result.
Worked Example: A Finite Endpoint Limit
Evaluate \(\displaystyle\lim_{x\to1^-}\frac{\pi/2-\arcsin x}{\sqrt{1-x}}\).
As \(x\to1^-\), the numerator tends to \(\pi/2-\pi/2=0\), and the denominator tends to zero through positive values. The quotient is defined for \(x<1\) near \(1\). Differentiate numerator and denominator on this open interval:
The cancellation is valid for \(x<1\), where \(1-x>0\). As a check, the original numerator and denominator are both positive on this side, and the final value \(\sqrt2\) is positive. The derivative ratio also approaches \(\sqrt2\), as required by the rule.
A Reliable Procedure
For an inverse-trigonometric quotient near a finite point, the following sequence keeps the domain and form conditions visible:
At an endpoint such as \(x=1\) for \(\arcsin x\), use \(x\to1^-\); at \(x=-1\), use \(x\to-1^+\).
Confirm that all functions in the quotient are defined for points sufficiently close to the limit point on that side.
Apply L’Hôpital’s rule only if the quotient has the required zero-over-zero form in the examples treated here.
Use the inverse-trigonometric derivative on the open approach interval, then simplify while respecting signs and domains.
Verify that the sign and finite or infinite behavior agree with the original quotient near the approach point.
Common Pitfall: Applying the Rule Without Checking the Form
Consider \(\displaystyle\lim_{x\to1^-}\frac{\arcsin x}{1-x}\). The numerator tends to \(\pi/2\), not zero, while the denominator tends to zero. This is not a zero-over-zero form, so applying L’Hôpital’s rule is not justified. In fact, the numerator remains positive and tends to a positive number, whereas the denominator is positive and tends to zero. Therefore the quotient tends to \(+\infty\).
A related error is to treat \(x\to1\) as a two-sided limit for an expression involving \(\arcsin x\). The function is not defined for real \(x>1\), so there is no two-sided real-domain approach through a full neighborhood of \(1\). The relevant limit is one-sided, \(x\to1^-\).
Why These Limits Matter
These quotients connect limit calculations to local rates of change. For example, \(\arcsin x/x\) compares the change in \(\arcsin x\) near zero with the change in its input. Its limit is the derivative of \(\arcsin\) at zero. At a domain endpoint, the same differentiation process can reveal more varied behavior: a quotient may diverge, or a denominator may balance the endpoint growth and produce a finite value.
The important lesson is not that every inverse-trigonometric quotient has the same answer. It is that the answer depends on the point, the approach side, the expression’s domain, and the form of the quotient. Check those conditions first, then differentiate only when the rule permits it.
Check Your Understanding
For each limit, first identify the approach side and the form of the quotient, then decide whether L’Hôpital’s rule applies.
- Evaluate \(\displaystyle\lim_{x\to0}\frac{\arcsin(3x)}{x}\).
- Evaluate \(\displaystyle\lim_{x\to0}\frac{\arctan x-x}{x^3}\).
- Evaluate \(\displaystyle\lim_{x\to-1^+}\frac{\arcsin x+\pi/2}{x+1}\). State whether the limit is finite or infinite.
- Explain why L’Hôpital’s rule cannot be applied directly to \(\displaystyle\lim_{x\to1^-}\frac{\arcsin x}{1-x}\).
- For \(\displaystyle\lim_{x\to1^-}\frac{\arccos x}{\sqrt{1-x}}\), verify the zero-over-zero form and determine the limit.
