Why an Indeterminate Product Needs Rewriting
A product can be difficult to evaluate when one factor approaches zero while another grows without bound. The factors pull in opposite directions: the first tends to shrink the product, while the second tends to enlarge it. The product’s limit depends on how quickly each factor changes.
For example, knowing only that \(u(x)\to 0\) and \(v(x)\to+\infty\) does not determine the limit of \(u(x)v(x)\). It could approach zero, a nonzero finite value, or infinity. The expression \(0\cdot\infty\) is therefore an indeterminate form, not an answer.
L’Hôpital’s Rule applies to certain quotients, not directly to products. The useful step is to rewrite the product as a quotient whose numerator and denominator both tend to zero or both grow without bound.
Two Quotient Orientations
Suppose \(u(x)\to 0\) and \(v(x)\to+\infty\) or \(-\infty\). Wherever the relevant reciprocals are defined, the product has either of these equivalent forms:
The first quotient has the form \(0/0\), because \(u(x)\to0\) and \(1/v(x)\to0\). The second has the form \(\infty/\infty\), because \(v(x)\) grows without bound and \(1/u(x)\) grows without bound in magnitude. The signs may be positive or negative; what matters is the quotient’s form and whether the limits and derivatives are defined as required.
These are identities, not approximations. For example, \(u/(1/v)=uv\) wherever \(v\ne0\), and \(v/(1/u)=uv\) wherever \(u\ne0\). Since a factor tending to positive or negative infinity is nonzero sufficiently near the limit point, its reciprocal is available on a sufficiently small punctured interval. For the other orientation, the factor tending to zero must also be nonzero on such an interval.
How the Conversion Makes L’Hôpital’s Rule Available
For a quotient \(F(x)/G(x)\) with the form \(0/0\) or \(\infty/\infty\), L’Hôpital’s Rule can be used under the theorem’s hypotheses. In particular, \(F\) and \(G\) must be differentiable on an interval near the limit point (apart from possibly at the point itself), \(G'(x)\) must be nonzero there, and the limit of \(F'(x)/G'(x)\) must exist, including as an infinite limit. Under those conditions, the quotient limit equals the derivative-quotient limit.
For a product, the conversion is what creates the quotient to which that result can apply. The reasoning is:
Confirm that one factor tends to zero and the other tends to positive or negative infinity.
Express it as \(u/(1/v)\) or \(v/(1/u)\), keeping the equality valid on a punctured interval.
Verify that it is \(0/0\) or \(\infty/\infty\) and that L’Hôpital’s Rule applies.
Find the limit of the derivative quotient, then use the rule to conclude the original product’s limit.
Worked Examples
Worked Example: A Logarithm Multiplied by a Vanishing Factor
Evaluate \(\displaystyle\lim_{x\to0^+}x\ln x\).
As \(x\to0^+\), \(x\to0\), while \(\ln x\to-\infty\). Rewrite the product in the orientation that gives a simple derivative quotient:
The numerator tends to \(-\infty\), and the denominator tends to \(+\infty\), so this is an infinity-over-infinity form. Differentiate the numerator and denominator separately:
Thus the derivative quotient is
As \(x\to0^+\), \(-x\to0\). L’Hôpital’s Rule therefore gives
Check the conversion: \(\ln x/(1/x)=x\ln x\) for \(x>0\), and the differentiated quotient simplifies to \(-x\), which does tend to zero.
Worked Example: A Trigonometric Product
Evaluate \(\displaystyle\lim_{x\to0}x\cot x\), where the expression is considered for \(x\ne0\) near zero.
Since \(\cot x=\cos x/\sin x\), it grows without bound in magnitude near zero, while \(x\to0\). Use the identity \(\cot x=1/\tan x\) to choose the quotient
Both numerator and denominator tend to zero. Their derivatives are
Because \(\sec^2x\ne0\) near zero, the derivative quotient is defined there. Applying L’Hôpital’s Rule gives
Hence \(\displaystyle\lim_{x\to0}x\cot x=1\). The check is that \(x/\tan x=x\cot x\) wherever \(\tan x\ne0\), and the derivative quotient simplifies to \(\cos^2x\), whose limit is one.
Worked Example: A Logarithm with a Squared Vanishing Factor
Evaluate \(\displaystyle\lim_{x\to0^+}x^2\ln(1/x)\).
Here \(x^2\to0\) and \(\ln(1/x)\to+\infty\). Choose the orientation
The numerator and denominator both tend to \(+\infty\). Their derivatives are
Therefore, the derivative quotient is
Its limit as \(x\to0^+\) is zero. L’Hôpital’s Rule gives
For a check, the quotient is exactly the original product for every \(x>0\), and the derivative quotient \(x^2/2\) approaches zero. The negative signs in the two derivatives cancel, so the quotient is positive, consistent with the original expression near zero.
Choosing an Orientation Efficiently
Both orientations are mathematically valid when their reciprocals are defined, but they need not be equally convenient. The goal is not to force a particular form; it is to make the derivative quotient manageable.
| Orientation | Typical quotient form | What to consider |
|---|---|---|
| \(u/(1/v)\) | \(0/0\) | Useful when differentiating \(u\) and \(1/v\) is straightforward. |
| \(v/(1/u)\) | \(\infty/\infty\) | Useful when differentiating \(v\) and \(1/u\) is more straightforward. |
For \(x\ln x\) near zero, \(\ln x/(1/x)\) leads to the simple derivative quotient \(-x\). The alternative \(x/(1/\ln x)\) is also an exact quotient on its domain, but differentiating \(1/\ln x\) produces a more complicated expression. Either form must still be checked against the rule’s hypotheses.
Why the Method Works
The conversion works because it preserves the product exactly while changing its algebraic form. If \(v(x)\ne0\), then dividing \(u(x)\) by \(1/v(x)\) multiplies \(u(x)\) by \(v(x)\). If \(u(x)\ne0\), then dividing \(v(x)\) by \(1/u(x)\) does the same. Thus no approximation has been made and no limit has been altered.
The important change is that L’Hôpital’s Rule concerns quotients with specific limiting forms. A product of the form \(0\cdot\infty\) does not meet that description, but its equivalent quotient can. Once a valid quotient is obtained, L’Hôpital’s Rule connects its limit to the limit of the derivatives’ quotient, provided the theorem’s hypotheses hold.
Common Mistake: Applying L’Hôpital Directly to a Product
A tempting but invalid calculation is to differentiate the two factors of \(x\ln x\) and multiply their derivatives:
This is wrong because L’Hôpital’s Rule does not say to differentiate factors of a product. In fact, the product tends to zero. The correction is to first write the product as a quotient with an indeterminate quotient form, such as \(\ln x/(1/x)\), then differentiate the quotient’s numerator and denominator separately under L’Hôpital’s Rule.
A second useful check is to avoid treating the symbols \(0\) and \(\infty\) as ordinary numbers. They describe factor behavior, not values that can be multiplied or canceled. The actual limit must be established using a valid rewrite and a justified limit method.
When to Pause Before Applying the Rule
The product-to-quotient step is appropriate only when the factors really have the stated limits. If a factor merely becomes large along some values but does not tend to positive or negative infinity, the expression may not have the required form. Also check that the chosen reciprocal is defined near the limit point and that the quotient is differentiable on the interval where the rule will be used.
After differentiating, simplify carefully. A correct rewrite does not guarantee an easy derivative quotient, and a complicated result is a reason to reconsider the orientation. Whichever orientation is chosen, the equality must hold on a punctured interval near the point, and the quotient’s form and derivative conditions must be checked before drawing a conclusion.
Check Your Understanding
For each question, identify the product form or quotient conversion before evaluating or explaining.
- As \(x\to0^+\), what quotient form results from rewriting \(x^3\ln x\) as \(\ln x/(1/x^3)\)?
- For \(x\to0^+\), rewrite \(x\ln(1/x)\) in both quotient orientations. Which orientation appears simpler to differentiate?
- Why is it invalid to apply L’Hôpital’s Rule directly to the product \(x\ln x\)?
- What conditions should be checked after converting a product to a quotient and before using L’Hôpital’s Rule?
- If \(u(x)\to0\) but \(u(x)\) is zero at additional points arbitrarily close to the limit point, what issue can arise when using the orientation \(v(x)/(1/u(x))\)?
