Why Infinite Endpoints Need Tail Conditions
When \(x\) tends to infinity, it does not approach a real number. Instead, it eventually lies beyond every fixed real number. Thus, applying L’Hôpital’s Rule at an infinite endpoint requires conditions that hold throughout an appropriate tail of the real line, not merely near one point.
For a limit as \(x\to+\infty\), a tail has the form \((A,\infty)\). For a limit as \(x\to-\infty\), a tail has the form \((-\infty,A)\). The functions must be differentiable there, and the quotient of their derivatives must have a limit as \(x\) moves along that tail toward the endpoint.
As in the finite-endpoint version of L’Hôpital’s Rule, the rule applies to a quotient only when direct substitution gives one of the two indeterminate forms \(0/0\) or \(\infty/\infty\). A quotient that merely looks difficult is not enough.
The Rule at Infinite Endpoints
The notation \(|f(x)|\to\infty\) includes functions tending to either positive or negative infinity. The theorem requires the derivative quotient to approach a finite limit; the examples here use that version of the rule.
In practice, organize the check in this order:
Evaluate the limiting behavior of numerator and denominator separately. The rule applies only to \(0/0\) or \(\infty/\infty\).
Confirm differentiability and that the denominator and its derivative are nonzero sufficiently far toward the endpoint.
Differentiate numerator and denominator separately, then evaluate the limit of \(f'(x)/g'(x)\).
When the theorem’s conditions hold, the derivative quotient’s limit is the original quotient’s limit.
Why the Rule Works: A Tail Argument
A short justification for the \(0/0\) case at \(+\infty\) shows why the derivative quotient must behave well throughout the tail. Suppose \(f(x)\to0\), \(g(x)\to0\), and \(f'(x)/g'(x)\to L\). Fix a sufficiently large \(x\) for which \(g(x)\ne0\). For any sufficiently large \(t>x\), the Cauchy Mean Value Theorem applied on \([x,t]\) gives a point \(\xi\in(x,t)\) such that
Since \(g(t)\to0\) and \(f(t)\to0\), the left side tends to \(f(x)/g(x)\) as \(t\to+\infty\). To see what this implies, let \(\varepsilon>0\). Since the derivative quotient tends to \(L\), choose a tail on which \(\left|f'(u)/g'(u)-L\right|<\varepsilon\) for every \(u\) on that tail. Choose \(x\) in this tail. Every point \(\xi\in(x,t)\) is also in the tail, so
Taking \(t\to+\infty\) preserves the resulting weak bound, giving \(\left|f(x)/g(x)-L\right|\le\varepsilon\). Because the tail can be chosen for every positive \(\varepsilon\), this proves \(f(x)/g(x)\to L\).
For the \(\infty/\infty\) case, apply the Cauchy Mean Value Theorem between a fixed point \(A\) in the tail and a variable point \(x>A\). If the derivative quotient stays within \(\varepsilon\) of \(L\) on that tail, then \((f(x)-f(A))/(g(x)-g(A))\) does too. As \(|g(x)|\to\infty\), the fixed terms \(f(A)\) and \(g(A)\) become negligible relative to \(g(x)\); the quotient of differences and \(f(x)/g(x)\) consequently differ by terms tending to zero. Letting \(\varepsilon\) be arbitrarily small gives the stated limit. For \(-\infty\), set \(t=-x\): differentiating \(f(-t)\) and \(g(-t)\) introduces the same factor \(-1\) in both derivatives, so their derivative quotient is unchanged.
Worked Examples
Worked Example: A Zero-Over-Zero Limit at Positive Infinity
Evaluate \(\lim_{x\to+\infty}\frac{\sin(1/x)}{1/x}\).
As \(x\to+\infty\), \(1/x\to0\), so \(\sin(1/x)\to0\) and \(1/x\to0\). This is the \(0/0\) form. Both functions are differentiable for \(x>0\), and the denominator \(1/x\) and its derivative \(-1/x^2\) are nonzero there.
Differentiate the numerator and denominator separately. By the chain rule,
Therefore, the derivative quotient is
L’Hôpital’s Rule gives the original limit as \(1\). As a direct check, let \(u=1/x\). Then \(u\to0^+\) and the original quotient is \(\sin u/u\), whose limit is \(1\).
Worked Example: An Infinity-Over-Infinity Limit at Positive Infinity
Evaluate \(\lim_{x\to+\infty}\frac{\ln x}{\sqrt{x}}\).
Both \(\ln x\) and \(\sqrt{x}\) tend to \(+\infty\), so this is an \(\infty/\infty\) form. For \(x>1\), both functions are differentiable, the denominator is nonzero, and its derivative \(1/(2\sqrt{x})\) is nonzero.
Taking the derivative quotient gives
Hence the original limit is \(0\). For an independent check, when \(x\ge1\), write \(\ln x=\int_1^x dt/t\). Since \(1/t\le t^{-3/4}\) for \(t\ge1\),
The lower and upper bounds both tend to \(0\), confirming the result.
Worked Example: A Zero-Over-Zero Limit at Negative Infinity
Evaluate \(\lim_{x\to-\infty}\frac{e^x}{xe^x}\).
As \(x\to-\infty\), \(e^x\to0\) and \(xe^x\to0\), so the form is \(0/0\). For \(x<-2\), the functions are differentiable, \(xe^x\ne0\), and its derivative \(e^x(x+1)\ne0\).
Differentiate the numerator and denominator:
Thus, on the tail \(x<-2\),
L’Hôpital’s Rule gives a limit of \(0\). Directly, the original quotient simplifies to \(1/x\), which also tends to \(0\) as \(x\to-\infty\).
Worked Example: An Infinity-Over-Infinity Limit at Negative Infinity
Evaluate \(\lim_{x\to-\infty}\frac{x^2}{-x^3}\).
As \(x\to-\infty\), the numerator \(x^2\to+\infty\) and the denominator \(-x^3\to+\infty\). On the tail \(x<-1\), both are differentiable, the denominator is nonzero, and its derivative \(-3x^2\) is nonzero.
The derivative quotient is
Therefore the original limit is \(0\). Checking directly, \(x^2/(-x^3)=-1/x\), and \(-1/x\to0\) as \(x\to-\infty\).
A Common Mistake: Using the Wrong Tail
The conditions must hold in the direction of the limit. For a limit as \(x\to-\infty\), checking behavior only for large positive \(x\) does not verify anything about the required tail. For instance, in the third example, \(e^x(x+1)\) is the derivative of the denominator, and it is nonzero for \(x<-2\). The relevant fact is its nonvanishing on that negative tail; its behavior for \(x>0\) is irrelevant.
Another error is to use the rule before checking the form. If a numerator tends to \(3\) and a denominator tends to \(0\), the quotient is not in the \(0/0\) or \(\infty/\infty\) form. Differentiating numerator and denominator in that situation is not justified by this rule.
Check Your Understanding
For each question, pay attention to both the limiting form and the direction of the tail.
- As \(x\to+\infty\), what indeterminate form does \(\ln x/x\) have?
- What must be checked about \(g'(x)\) before applying the rule to \(f(x)/g(x)\) on a tail?
- For \(x\to-\infty\), does checking the derivative conditions only for large positive \(x\) suffice? Explain briefly.
- For \(x\to-\infty\), compute the derivative quotient for \(f(x)=x\) and \(g(x)=x^2\), and state the limit of the original quotient.
- If \(f(x)\to 2\) and \(g(x)\to0\), is this one of the indeterminate forms covered by this version of L’Hôpital’s Rule?
