Why a Product Needs Rewriting
A product whose factors tend to zero and infinity has no determined limit from those facts alone. The factor tending to zero may dominate, the factor tending to infinity may dominate, or their rates may balance. The notation \(0\cdot\infty\) names this uncertainty; it is not an arithmetic operation that determines an answer.
A useful next step is to rewrite the product as a quotient. The quotient may then have one of the forms \(0/0\) or \(\infty/\infty\), for which L'Hôpital's Rule may apply, provided its hypotheses are satisfied. The algebraic rewrite is exact; it does not, by itself, establish the limit.
Two Quotient Conversions
Suppose \(f(x)\to 0\) and \(|g(x)|\to\infty\) as \(x\) approaches an endpoint. Wherever \(g(x)\ne 0\), the product can be written in either of these ways:
In the first quotient, \(1/g(x)\to 0\), so its numerator and denominator both tend to zero. This gives a \(0/0\) form. In the second, \(1/f(x)\) has magnitude tending to infinity, so both numerator and denominator have magnitude tending to infinity. This gives an \(\infty/\infty\) form.
These conversions require the reciprocal expressions to be defined sufficiently close to the endpoint. For example, \(g(x)\) must be nonzero there to use \(1/g(x)\), and \(f(x)\) must be nonzero there to use \(1/f(x)\). It is enough that the needed function be nonzero on a punctured interval or a suitable one-sided neighborhood; its value at the endpoint itself is not required.
The choice is often practical rather than deep. Use the form whose derivatives are simpler and whose denominator derivative satisfies the conditions of L'Hôpital's Rule. The rule applies to a quotient of type \(0/0\) or \(\infty/\infty\) when the functions are differentiable near the endpoint, the denominator derivative is nonzero there, and the derivative quotient has a limit (finite or infinite). A conclusion about the original product follows because the quotient equals that product wherever the expressions are defined.
Why the Conversion Is Valid
The key justification is the reciprocal limit. If \(|g(x)|\to\infty\), then \(1/g(x)\to 0\), wherever \(g\) is nonzero. Indeed, for any \(\varepsilon>0\), the condition \(|g(x)|>1/\varepsilon\) implies \(|1/g(x)|<\varepsilon\). Thus, if also \(f(x)\to0\), the quotient \(f(x)/(1/g(x))\) has the required \(0/0\) limiting form. The identity between that quotient and the product is ordinary algebra.
Here is the conversion as a procedure:
Record which factor tends to zero and which has magnitude tending to infinity, including the direction of approach and any sign information.
Rewrite the product as either \(f/(1/g)\), a \(0/0\) form, or \(g/(1/f)\), an \(\infty/\infty\) form, ensuring the reciprocal is defined nearby.
Confirm the quotient has the required form and that L'Hôpital's Rule applies on the relevant neighborhood.
Find the limit of the derivative quotient. If it exists, L'Hôpital's Rule gives the limit of the quotient, and hence of the original product.
Worked Logarithmic Limits
Worked Example: \(x\ln x\) as \(x\to0^+\)
As \(x\to0^+\), \(x\to0\) and \(\ln x\to-\infty\), so the product has the form \(0\cdot\infty\), with a negative sign. Rewrite it as a quotient:
Here \(\ln x\to-\infty\) and \(1/x\to+\infty\), so this is an \(\infty/\infty\) form in magnitude. For \(x>0\), both functions are differentiable and the derivative of the denominator is \(-1/x^2\), which is nonzero. Apply L'Hôpital's Rule:
Therefore \(\lim_{x\to0^+}x\ln x=0\). The product approaches zero from below because \(x>0\) and \(\ln x<0\) for \(0<x<1\). This sign is consistent with the limit. The derivative quotient also simplifies exactly to \(-x\), whose limit is zero.
Worked Example: \((1/x)\ln x\) as \(x\to\infty\)
Now the product has a factor \(1/x\to0\) and a factor \(\ln x\to+\infty\). A convenient conversion gives an \(\infty/\infty\) quotient directly:
For \(x>1\), numerator and denominator tend to \(+\infty\), and the denominator derivative is \(1\), which is nonzero. Both functions are differentiable for \(x>1\). L'Hôpital's Rule yields
Hence \(\lim_{x\to\infty}(1/x)\ln x=0\). The derivative quotient \(1/x\) tends to zero, so the value obtained from the rule is consistent with the converted quotient and the original product.
An Exponential Example
Worked Example: \(xe^{-x}\) as \(x\to\infty\)
Here \(x\to+\infty\), while \(e^{-x}\to0\). Write the product as a quotient of type \(\infty/\infty\):
The identity follows from \(e^{-x}=1/e^x\), and the denominator is positive. The quotient has numerator and denominator tending to \(+\infty\). Both are differentiable, and the derivative of the denominator, \(e^x\), is never zero. L'Hôpital's Rule gives
Thus \(\lim_{x\to\infty}xe^{-x}=0\). As a check, the converted quotient is exactly the original product, and the final expression \(e^{-x}\) tends to zero.
Conditions and Signs Matter
A product's sign does not decide whether its limit is zero, finite and nonzero, or infinite. It does help check the conclusion. For example, the first worked example is negative near zero, so a positive limit would have signaled an error. In contrast, \(xe^{-x}\) is positive, consistent with its limit of zero.
Also, the phrase “apply L'Hôpital” does not replace checking the quotient's form or the hypotheses. A product cannot be passed directly to the rule: the rule concerns a quotient, not a product. After conversion, the numerator and denominator must have the appropriate limiting form. The functions must be differentiable on the relevant punctured neighborhood, and the denominator derivative must not vanish there. The behavior of the derivative quotient must then support the claimed conclusion.
Common Mistake: Differentiating the Product Factors
A tempting but incorrect approach to \(x\ln x\) is to differentiate both factors separately and treat the product's limit as the limit of a quotient of derivatives. For instance, one might write
The equality marked wrong is not justified: L'Hôpital's Rule does not say that the limit of a product equals a quotient of derivatives. In this particular example the final answer happens to be correct, but that does not validate the calculation. The correct route is to first write \(x\ln x=\ln x/(1/x)\), verify the \(\infty/\infty\) form and the rule's conditions, and then differentiate the numerator and denominator. This gives the derivative quotient \((1/x)/(-1/x^2)=-x\), whose limit is zero.
Why This Method Is Useful
Rewriting a product exposes the relative rates of its factors in a form suited to comparison. A logarithm may grow without bound while a reciprocal tends to zero; an exponential reciprocal may shrink while a polynomial factor grows. The quotient lets L'Hôpital's Rule compare those rates through derivatives.
There is no single conversion that is always best. For a product \(f(x)g(x)\), the quotient \(f(x)/(1/g(x))\) may produce a simpler derivative ratio than \(g(x)/(1/f(x))\), or the reverse may be true. Both are legitimate when their reciprocals are defined near the endpoint. Choose deliberately, keep track of signs, and check the hypotheses before differentiating.
Check Your Understanding
For each question, identify the limiting form and explain the conversion or condition involved.
- As \(x\to0^+\), what are the separate limits of \(x\) and \(\ln x\) in the product \(x\ln x\)?
- Rewrite \(t e^{-t}\), as \(t\to\infty\), as a quotient of type \(\infty/\infty\).
- For a product \(f(x)g(x)\) with \(f(x)\to0\) and \(|g(x)|\to\infty\), what limits does \(f(x)/(1/g(x))\) have in its numerator and denominator?
- Why is it invalid to apply L'Hôpital's Rule directly to a product?
- What must be checked about the denominator derivative before applying L'Hôpital's Rule?
