What “Indeterminate” Means
L’Hopital’s Rule applies directly to two specific patterns: \(0/0\) and \(\infty/\infty\). But limits can also involve products, differences, and powers whose factors have competing behavior. The important first step is to recognize that these patterns do not determine a limit by themselves. They indicate that more analysis is needed.
For example, if \(f(x)\to0\) and \(g(x)\to\infty\), the product \(f(x)g(x)\) might tend to zero, a nonzero finite value, or infinity. The particular functions matter. Calling the pattern “indeterminate” does not mean that the limit fails to exist; it means that the information in the pattern alone is insufficient to decide what the limit is.
Here, \(\infty\) describes unbounded growth in the relevant direction; in the common positive-infinity versions of these forms, the functions tend to \(+\infty\). The quotient forms \(0/0\) and \(\infty/\infty\) are the ones to which L’Hopital’s Rule applies directly, provided its other hypotheses also hold. The remaining forms often need to be rewritten first.
The Same Form Can Lead to Different Limits
A useful way to understand indeterminacy is to compare expressions with the same component behavior but different outcomes. The examples below do not yet call for L’Hopital’s Rule; they show why identifying a form is only the beginning of the work.
Worked Example: The Form \(0\cdot\infty\)
As \(x\to0^+\), the factor \(x\) tends to zero and the factor \(1/x\) tends to infinity, but their product is exactly one:
Changing the second factor changes the outcome, even though it still tends to infinity. For example,
Thus the same \(0\cdot\infty\) pattern can produce the limits \(0\), \(1\), and \(2\). Neither factor’s limit alone determines the product.
Worked Example: The Form \(\infty-\infty\)
As \(x\to\infty\), both \(x+1\) and \(x\) tend to infinity, but their difference is constant:
In contrast, \(2x-x=x\to\infty\), while \(x-2x=-x\to-\infty\). All three expressions subtract quantities that tend to infinity. The outcomes differ because the size of the difference depends on how the two terms relate to one another.
The same lesson applies to the power forms. For instance, as \(x\to\infty\), the base \(1+1/x\) tends to \(1\), while the exponent \(x\) tends to infinity. Yet changing the exponent can change the limit:
These familiar limits illustrate the \(1^\infty\) form. The base approaches \(1\), but the small difference between the base and \(1\) can accumulate under a growing exponent. Similar competition occurs in \(0^0\) and \(\infty^0\); taking logarithms makes that competition visible.
Turning a Product into a Quotient
Suppose \(f(x)\to0\) and \(|g(x)|\to\infty\). Since \(1/g(x)\to0\), the product can be written as a quotient of two quantities tending to zero. This does not automatically solve the limit, but it can put the expression into the \(0/0\) form required for L’Hopital’s Rule.
Proof. Since \(|g(x)|\to\infty\), there is a region sufficiently near the limiting endpoint on which \(|g(x)|>1\). In particular, \(g(x)\ne0\) there, so the quotient is defined. Also,
On that region, multiplying the numerator and denominator of \(f(x)/(1/g(x))\) gives
The numerator \(f(x)\) and denominator \(1/g(x)\) both tend to zero, as claimed. \(\square\)
When the transformed quotient is differentiable and the other hypotheses of the zero-over-zero form of L’Hopital’s Rule hold, the rule may then be used. The conversion establishes the form; it does not establish those additional hypotheses.
Worked Example: Rewriting a Product Before Using L’Hopital’s Rule
Consider \(x\ln x\) as \(x\to0^+\). Here \(x\to0\) and \(\ln x\to-\infty\), so the product has the \(0\cdot\infty\) pattern. Rewrite it as
Both numerator and denominator tend to \(-\infty\) and \(+\infty\), respectively, so this is an \(\infty/\infty\) form in absolute magnitude. For \(x>0\), the denominator derivative is \(-1/x^2\), which is nonzero. The quotient of derivatives is
Set \(s=1/x\), so \(x\ln x=-\dfrac{\ln s}{s}\) as \(s\to+\infty\). The quotient of derivatives is \(-1/s\to0\), so the infinity-over-infinity form of L’Hopital’s Rule gives
The sign is consistent with the result: \(\ln x<0\) for \(0<x<1\), so \(x\ln x\) approaches zero through negative values.
Resolving Some \(\infty-\infty\) Forms
There is no single algebraic transformation that resolves every difference of two terms tending to infinity. A useful technique is to combine the terms in a way that exposes their cancellation. For differences involving square roots, multiplying by the conjugate often converts subtraction into a quotient with a simpler numerator.
Worked Example: Rationalizing a Difference of Square Roots
Evaluate
Each term inside the difference tends to \(+\infty\), so direct substitution gives the \(\infty-\infty\) form. For \(x>0\), multiply and divide by the conjugate:
Dividing the numerator and denominator by \(x\), which is positive, yields
As \(x\to\infty\), \(3/x\to0\), so continuity of the square-root function gives
The conjugate exposes the difference between the squared quantities: \((x^2+3x)-x^2=3x\). It turns the original cancellation problem into a quotient whose limit is directly accessible.
Taking Logarithms of Indeterminate Powers
For a power \(u(x)^{v(x)}\), logarithms convert multiplication in the exponent into an ordinary product:
provided \(u(x)>0\). This identity explains why the power forms \(1^\infty\), \(0^0\), and \(\infty^0\) are usually analyzed by studying \(v(x)\ln u(x)\). The transformed expression may be a product or quotient that can then be estimated or treated with L’Hopital’s Rule when its hypotheses are satisfied.
Proof. Positivity of \(u(x)\) allows us to take its logarithm. The standard logarithm and exponential identity gives, at every point under consideration,
If \(M\) is finite, continuity of the exponential function implies that the right-hand side tends to \(\exp(M)\). If \(M=+\infty\), then for every real \(B\), the quantity \(v(x)\ln u(x)\) is eventually greater than \(\ln B\) when \(B>0\). Since the exponential function is increasing, its exponential is eventually greater than \(B\), which proves divergence to \(+\infty\). If \(M=-\infty\), then for every \(\varepsilon>0\), the quantity \(v(x)\ln u(x)\) is eventually less than \(\ln\varepsilon\). Exponentiating gives \(u(x)^{v(x)}<\varepsilon\); the power is positive, so it tends to zero. This proves all three cases. \(\square\)
Worked Example: Resolving a \(0^0\) Form
Find the limit of \(x^x\) as \(x\to0^+\). The base tends to zero and the exponent tends to zero, so the expression has the \(0^0\) form. For every \(x>0\),
Rewrite the exponent as a quotient:
The numerator tends to \(-\infty\), and the denominator tends to \(+\infty\). Differentiating numerator and denominator gives
Set \(s=1/x\), so \(x\ln x=-\dfrac{\ln s}{s}\) as \(s\to+\infty\). The quotient of derivatives is \(-1/s\to0\), so the infinity-over-infinity form of L’Hopital’s Rule shows that \(x\ln x\to0\). The logarithmic reduction theorem now gives
This illustrates why \(0^0\) cannot be evaluated from the base and exponent limits alone: the limiting value depends on the rate at which each changes.
Worked Example: Resolving a \(1^\infty\) Form
Consider
The base tends to \(1\) and the exponent tends to \(+\infty\). Set
Both numerator and denominator tend to zero. For \(x>0\), the denominator derivative is \(-1/x^2\), and the numerator derivative is
Thus the quotient of derivatives is
L’Hopital’s Rule gives \(L(x)\to2\). Applying the logarithmic reduction theorem,
The logarithm reveals the accumulated effect of the base’s small deviation from \(1\).
Choosing a Transformation Carefully
Recognizing a form does not license a particular operation automatically. The transformation must preserve the expression on a region where it is defined, and the resulting limit must meet the hypotheses of any theorem used. In particular, L’Hopital’s Rule is not a general instruction to differentiate whenever a limit is difficult.
- For \(0/0\) and \(\infty/\infty\): check the differentiability and nonzero-derivative conditions before applying L’Hopital’s Rule.
- For \(0\cdot\infty\): rewrite the product as a quotient, then check the form and the hypotheses of the rule.
- For \(\infty-\infty\): look for cancellation, a common denominator, or a conjugate; verify that the rewritten expression is equal to the original on the relevant domain.
- For \(1^\infty\), \(0^0\), and \(\infty^0\): make sure the base is positive before taking logarithms, then study the exponent times the logarithm of the base.
The key distinction is between a form and a conclusion. A form describes the component behavior; it does not supply the missing information about their relative rates. Algebraic rewriting, logarithms, and L’Hopital’s Rule are tools for uncovering those rates, each with its own conditions.
Check Your Understanding
Use the definitions, transformations, and examples in this tutorial to answer the following questions.
- Why does the pattern \(0\cdot\infty\) not determine the limit of a product?
- How can a product whose factors tend to zero and infinity be rewritten as a quotient of two quantities tending to zero?
- What algebraic step resolves the difference \(\sqrt{x^2+3x}-x\)?
- For a positive base \(u(x)\), which expression should be studied to analyze \(u(x)^{v(x)}\)?
- Why must the hypotheses of L’Hopital’s Rule still be checked after an indeterminate form has been rewritten?