When L'Hôpital's Rule Produces an Infinite Limit

When the Derivative Ratio Diverges

A quotient in an indeterminate form can itself tend to positive or negative infinity. In this situation, L’Hôpital’s Rule can identify not just that the quotient is unbounded, but its direction of divergence.

The key condition is stronger than simply having an indeterminate form: the quotient of derivatives must tend to positive or negative infinity. The sign matters. A derivative ratio tending to positive infinity leads to a quotient tending to positive infinity under the hypotheses below; a derivative ratio tending to negative infinity leads to a quotient tending to negative infinity.

Precise Conditions

First consider a finite endpoint \(a\), approached from one side. Suppose \(f\) and \(g\) are continuous at \(a\), differentiable on the punctured interval on the side of approach, and \(f(a)=g(a)=0\). Assume also that \(g(x)\ne 0\) and \(g'(x)\ne 0\) throughout a sufficiently small punctured interval on that side. If

then \(\lim_{x\to a} f(x)/g(x)=+\infty\). If the derivative ratio tends to \(-\infty\), then the quotient tends to \(-\infty\).

For an endpoint at infinity, take \(x\to+\infty\). Suppose \(f\) and \(g\) are differentiable on a ray, \(|f(x)|\to\infty\), \(|g(x)|\to\infty\), and \(g(x)\ne 0\) and \(g'(x)\ne 0\) for all sufficiently large \(x\). If the derivative ratio tends to \(+\infty\), then \(f(x)/g(x)\to+\infty\); if it tends to \(-\infty\), then \(f(x)/g(x)\to-\infty\). The same result holds as \(x\to-\infty\), with a ray extending toward negative infinity.

In each case, \(\lim f/g\) is an infinite conclusion, not an ordinary finite-limit calculation. The assumptions ensure that the quotient is defined near the endpoint and that the derivative ratio can be used there. In particular, the denominator derivative must not vanish nearby.

Why the Finite-Endpoint Result Works

The Cauchy Mean Value Theorem gives the essential link. For \(x\ne a\) near \(a\), apply it to \(f\) and \(g\) on the interval with endpoints \(a\) and \(x\). Continuity at the endpoints and differentiability between them are satisfied. Since \(f(a)=g(a)=0\), there is a point \(c\) strictly between \(a\) and \(x\) such that

As \(x\) approaches \(a\), every point \(c\) between \(a\) and \(x\) also approaches \(a\). Thus, if \(f'(t)/g'(t)\) exceeds any prescribed positive bound near \(a\), then \(f'(c)/g'(c)\) exceeds that bound for \(x\) sufficiently close to \(a\). This is exactly the definition of the quotient tending to \(+\infty\). The same reasoning with bounds below any prescribed negative threshold proves the \(-\infty\) case.

Why the Infinity-Endpoint Result Needs a Different Argument

At \(+\infty\), the quotient \(f(x)/g(x)\) is not a \(0/0\) quotient, so the preceding endpoint argument does not apply. In particular, if the Cauchy Mean Value Theorem is applied between a fixed point and \(x\), the resulting point \(c\) need not approach infinity. Instead, choose the fixed point far enough out that the derivative ratio has the required bound everywhere beyond it.

Here is the proof when \(f'(x)/g'(x)\to+\infty\). Let \(K>0\) be any desired lower bound for \(f(x)/g(x)\). Since the derivative ratio tends to \(+\infty\), choose \(B\) so large that

Because \(|g(x)|\to\infty\), for all sufficiently large \(x\), \(g(x)\ne g(B)\), and

Apply the Cauchy Mean Value Theorem to \(f\) and \(g\) on \([B,x]\). The point \(c\) it supplies satisfies \(B<c<x\), so it lies in the region where the derivative-ratio bound holds. Therefore, writing \(Q=(f(x)-f(B))/(g(x)-g(B))\), we have \(Q=f'(c)/g'(c)>2(K+1)\). Now decompose the original quotient:

Since this holds for every \(K>0\) once \(x\) is sufficiently large, \(f(x)/g(x)\to+\infty\). For the negative case, choose \(B\) so that \(f'(t)/g'(t)<-2(K+1)\) for \(t>B\). The same positive factor \((g(x)-g(B))/g(x)>1/2\), together with \(|f(B)/g(x)|<1\), gives \(f(x)/g(x)<-K\). This proves the \(-\infty\) conclusion.

Worked Example: A Positive Infinite Limit from a \(0/0\) Form

Worked Example: A Negative Infinite Limit from a \(0/0\) Form

Worked Example: Positive Divergence at Infinity

Worked Example: Negative Divergence at Infinity

A Common Sign Mistake

A common error is to treat “infinite” as though it always meant positive infinity. For the second example, one might write

This loses the sign. In fact, the derivative ratio \(e^x/(2x)\) tends to \(-\infty\) as \(x\to0^-\), not \(+\infty\). The corrected conclusion is \(-\infty\). Always determine whether the derivative ratio becomes arbitrarily large positive or arbitrarily large negative; then check that the original quotient’s sign is consistent.

When This Result Is Useful

This form of L’Hôpital’s Rule is useful when the quotient is initially an indeterminate \(0/0\) or \(\infty/\infty\) form and direct simplification is not apparent. It is especially important to distinguish the direction of divergence: the results \(+\infty\) and \(-\infty\) are not interchangeable, and signs often reveal an error immediately.

The hypotheses are not optional. The original quotient must be defined near the limit point, the denominator derivative must be nonzero nearby, and the derivative ratio must actually tend to a signed infinity. A derivative ratio that merely changes sign or is unbounded without tending to one signed infinity does not meet this condition. At infinity, the proof also depends on choosing a fixed point \(B\) far enough along the ray so that the mean-value point lies where the derivative-ratio bound holds.