Why Nested Logarithms Need Care
A logarithm grows without bound as its argument grows without bound, but it does so slowly. Applying a logarithm again slows the growth further: for sufficiently large \(x\), both \(\ln x\) and \(\ln(\ln x)\) increase without bound, yet \(\ln(\ln x)\) grows much more slowly. Limits comparing these quantities give a precise way to express that difference in growth.
The main example is the quotient \(\frac{\ln(\ln x)}{\ln x}\) as \(x\) tends to infinity. Its numerator and denominator both tend to positive infinity. That makes it an infinity-over-infinity indeterminate form, so L’Hôpital’s Rule may be relevant. The key is to verify the form and the rule’s conditions before differentiating, then apply the chain rule carefully.
When L’Hôpital’s Rule Applies at Infinity
For the limits here, the relevant version of L’Hôpital’s Rule has the following form. Suppose \(f\) and \(g\) are differentiable for all sufficiently large \(x\), \(g'(x)\ne 0\) there, and both \(f(x)\) and \(g(x)\) tend to positive infinity. If the derivative quotient has a limit \(L\), then the original quotient has the same limit:
The hypotheses matter. The original quotient must first have the required indeterminate form; the functions must be differentiable on a sufficiently large interval; the denominator’s derivative must not vanish there; and the derivative quotient must have a limit. A quotient that merely looks complicated is not, for that reason alone, eligible for this rule.
Why does the derivative quotient provide information about the original one? The Cauchy Mean Value Theorem explains the connection. Fix a sufficiently large \(X\). For any \(x>X\), the theorem gives a point \(c\) between \(X\) and \(x\) such that
If the derivative quotient is close to \(L\) for every sufficiently large input, then the quotient of these increments is close to \(L\) once \(X\) is chosen sufficiently large. As \(x\) tends to infinity, \(g(x)\) tends to infinity, so the fixed starting values \(f(X)\) and \(g(X)\) become negligible relative to \(f(x)\) and \(g(x)\). Thus the quotient of the functions has the same limit as the quotient of their derivatives. This is the reason for the rule, not a license to differentiate any quotient at will.
The Basic Nested-Logarithm Limit
Worked Example: Comparing \(\ln(\ln x)\) with \(\ln x\)
Evaluate \(\lim_{x\to\infty}\frac{\ln(\ln x)}{\ln x}\).
Step 1: Check the form and domain. For \(x>1\), \(\ln x>0\), so \(\ln(\ln x)\) is defined. As \(x\to\infty\), \(\ln x\to\infty\), and consequently \(\ln(\ln x)\to\infty\). The denominator also tends to infinity. This is an infinity-over-infinity form.
Step 2: Differentiate numerator and denominator. By the chain rule, the derivative of \(\ln(\ln x)\) is the derivative of the outer logarithm, \(1/\ln x\), multiplied by the derivative of its inner function, \(1/x\). The denominator has derivative \(1/x\):
Step 3: Apply L’Hôpital’s Rule. The derivatives exist for \(x>1\), and the denominator derivative \(1/x\) is nonzero there. The derivative quotient is
Since \(\ln x\to\infty\), this derivative quotient tends to \(0\). Therefore, L’Hôpital’s Rule gives
Check. Set \(t=\ln x\). As \(x\to\infty\), \(t\to\infty\), and the original quotient becomes \(\frac{\ln t}{t}\). The derivative calculation above establishes that this quotient tends to \(0\), consistent with the result. The two original expressions both tend to infinity, and all the required conditions for this application of the rule hold.
Comparing Three Levels of Logarithm
The same method can compare a logarithm nested three times with one nested twice. The chain rule accumulates one factor for each inner logarithm, so it is especially important to write the derivative in stages rather than omit a factor.
Worked Example: Comparing \(\ln(\ln(\ln x))\) with \(\ln(\ln x)\)
Evaluate \(\lim_{x\to\infty}\frac{\ln(\ln(\ln x))}{\ln(\ln x)}\).
Step 1: Check the form and domain. For \(x>e\), \(\ln(\ln x)>0\), so the numerator is defined. As \(x\to\infty\), \(\ln(\ln x)\to\infty\), and hence \(\ln(\ln(\ln x))\to\infty\). The denominator also tends to infinity.
Step 2: Compute both derivatives. Differentiating the numerator from the outside inward gives
The denominator’s derivative is
Step 3: Apply the rule and evaluate the derivative quotient. The denominator derivative is nonzero for \(x>e\), and the functions are differentiable there. Thus
Both original functions tend to positive infinity, so L’Hôpital’s Rule applies and the requested limit is \(0\).
Check. Let \(t=\ln(\ln x)\). Then \(t\to\infty\), and the original quotient is \(\frac{\ln t}{t}\), which is the same growth comparison as in the basic example. Its limit is therefore consistent with \(0\). The derivative quotient also confirms directly that the extra nesting contributes a factor \(1/\ln(\ln x)\), which tends to zero.
When Similar Nested Expressions Have a Different Limit
Not every quotient of nested logarithms tends to zero. If the numerator and denominator differ only by a constant inside one logarithmic argument, their ratio can instead tend to one. L’Hôpital’s Rule still works, provided the form and hypotheses are checked.
Worked Example: A Nested-Logarithm Quotient Tending to One
Evaluate \(\lim_{x\to\infty}\frac{\ln(\ln x)}{\ln(\ln(x^2))}\).
Step 1: Check the form. For \(x>1\), both expressions are defined. The numerator tends to infinity. Since \(\ln(x^2)=2\ln x\to\infty\), the denominator \(\ln(\ln(x^2))\) also tends to infinity.
Step 2: Differentiate. The numerator derivative is \(1/(x\ln x)\). For the denominator, apply the chain rule to the outer logarithm and then differentiate \(\ln(x^2)\):
The denominator derivative is nonzero for \(x>1\). The derivative quotient is therefore
L’Hôpital’s Rule applies to the infinity-over-infinity form, giving a limit of \(1\).
Check. Since \(x>1\), \(\ln(x^2)=2\ln x\), and therefore
Thus the original quotient is \(\frac{\ln(\ln x)}{\ln 2+\ln(\ln x)}\). Dividing its numerator and denominator by \(\ln(\ln x)\), which tends to infinity, gives \(\frac{1}{\frac{\ln 2}{\ln(\ln x)}+1}\to 1\). This exact identity independently confirms the answer.
A Common Chain-Rule Mistake
A frequent error is to differentiate \(\ln(\ln x)\) as though its derivative were just \(1/\ln x\). That expression accounts for differentiating the outer logarithm but leaves out the derivative of the inner function \(\ln x\), which is \(1/x\).
The correction is to differentiate each layer in turn:
Consequently, when the denominator is \(\ln x\), the correct derivative quotient is \(\frac{1/(x\ln x)}{1/x}=\frac{1}{\ln x}\), not \(\frac{x}{\ln x}\). The chain rule is not an optional refinement: omitting an inner derivative changes the limit problem itself.
A Reliable Procedure
For a quotient involving nested logarithms at infinity, use a deliberate sequence of checks. Each step protects against a different error.
Make sure every logarithm has a positive argument for all sufficiently large \(x\).
Verify that both numerator and denominator tend to positive infinity before using the infinity-over-infinity version of L’Hôpital’s Rule.
For each nested logarithm, include the derivative of its inner function.
Confirm differentiability and that the denominator derivative is nonzero for sufficiently large \(x\).
Use its limit to determine the original limit, then check the result against an exact identity or a simpler growth comparison when possible.
The central comparison is that adding another logarithm can make a quantity grow far more slowly than the expression it is nested inside. L’Hôpital’s Rule makes that comparison precise by reducing the quotient to a derivative quotient. The result depends on correct chain-rule differentiation and on satisfying the rule’s hypotheses—not on the appearance of the original expressions alone.
Check Your Understanding
For each question, identify the form and explain which facts justify the limit method you choose.
- As \(x\to\infty\), what form does \(\frac{\ln(\ln x)}{\ln x}\) have, and what are the derivatives of its numerator and denominator?
- Why must the derivative of \(\ln(\ln x)\) include a factor of \(1/x\)?
- In comparing \(\ln(\ln(\ln x))\) with \(\ln(\ln x)\), what is the derivative quotient?
- Use \(\ln(x^2)=2\ln x\) to explain why \(\frac{\ln(\ln x)}{\ln(\ln(x^2))}\) tends to \(1\).
- Name two conditions, besides an infinity-over-infinity form, that must be checked before applying L’Hôpital’s Rule.
