Why a Difference of Diverging Logarithms Needs Care
Suppose two logarithmic expressions both grow without bound as \(x\) increases. Their difference may nevertheless approach zero, approach a finite nonzero value, grow without bound, or fail to have a limit. Knowing that each term tends to infinity does not determine what their difference does.
The useful first move is often to combine the logarithms. A difference such as \(\ln A(x)-\ln B(x)\) can be rewritten as a single logarithm involving the quotient \(A(x)/B(x)\), provided both arguments are positive. The problem then becomes one of finding the limit of that quotient.
Sometimes the quotient can be simplified directly. If it is still an indeterminate infinity-over-infinity form, L’Hôpital’s Rule may apply to the quotient. The rule applies to the quotient—not directly to the original difference.
Combine the Logarithms First
The positivity condition matters: the real natural logarithm is defined only for positive arguments. When the expressions inside the logarithms are positive for all sufficiently large \(x\), the identity applies throughout the part of the domain relevant to a limit as \(x\to\infty\).
The identity follows from the logarithm law \(\ln u-\ln v=\ln u+\ln(1/v)\). Since \(u>0\) and \(v>0\), we have \(1/v>0\), so the product law gives \(\ln u+\ln(1/v)=\ln(u/v)\). Taking \(u=A(x)\) and \(v=B(x)\) proves the stated identity.
This rewriting also explains what determines the limit. If \(A(x)/B(x)\) tends to a positive number \(r\), continuity of \(\ln\) at \(r\) gives \(\ln(A(x)/B(x))\to\ln r\). If the quotient tends to infinity, its logarithm tends to infinity; if the quotient tends to zero through positive values, its logarithm tends to negative infinity.
When L’Hôpital’s Rule Can Help
After combining the logarithms, the quotient \(A(x)/B(x)\) may still be difficult to evaluate. If both \(A(x)\) and \(B(x)\) tend to infinity, then this quotient has the indeterminate form infinity over infinity. Under the usual conditions for L’Hôpital’s Rule, the quotient’s limit can be found by studying \(A'(x)/B'(x)\).
For an infinite endpoint, the relevant conditions are that \(A\) and \(B\) are differentiable for all sufficiently large \(x\), that \(B'(x)\ne 0\) there, and that both \(A(x)\) and \(B(x)\) tend to infinity. If \(A'(x)/B'(x)\) has a limit (finite or infinite), L’Hôpital’s Rule gives the same limit for \(A(x)/B(x)\). After finding the quotient limit, return to the logarithm and use continuity or the appropriate infinite-endpoint behavior of \(\ln\).
It is worth checking the quotient before differentiating. Factoring, dividing by a dominant power, or using a logarithm identity may make the limit immediate. L’Hôpital’s Rule is not a replacement for simplification; it is a tool for a quotient that remains indeterminate.
Worked Example: A Difference That Tends to Zero
Evaluate \(\displaystyle\lim_{x\to\infty}\bigl(\ln(x+3)-\ln(x+1)\bigr)\).
For \(x\) sufficiently large, both arguments are positive and both logarithms tend to infinity. Combine them:
The quotient inside the logarithm is an infinity-over-infinity form. Its numerator and denominator are differentiable, and the derivative of the denominator is \(1\), which is nonzero. L’Hôpital’s Rule gives
Since \(\ln\) is continuous at \(1\),
Thus the original difference tends to \(0\), even though each logarithm separately tends to infinity. As a check, the combined expression is \(\ln(1+2/(x+1))\); its argument tends to \(1\), confirming the same limit.
A Quotient That Grows Without Bound
The quotient of the arguments does not always approach a positive finite number. If it grows without bound, the logarithm of that quotient does too, so the difference of logarithms tends to infinity. L’Hôpital’s Rule can establish this when the quotient is an indeterminate infinity-over-infinity form and its derivative quotient tends to infinity.
Worked Example: A Difference That Tends to Infinity
Evaluate \(\displaystyle\lim_{x\to\infty}\bigl(\ln(x^2+4x+1)-\ln(x+1)\bigr)\).
For sufficiently large \(x\), both arguments are positive and tend to infinity. Combining the logarithms gives
The quotient is of type infinity over infinity. Its numerator and denominator are differentiable, and the denominator derivative is \(1\), so L’Hôpital’s Rule applies:
Because the quotient is positive for sufficiently large \(x\) and tends to infinity, its natural logarithm also tends to infinity. Therefore,
To check against the original expression, polynomial division gives \(\frac{x^2+4x+1}{x+1}=x+3-\frac{2}{x+1}\), which tends to infinity. Taking its logarithm therefore confirms the result.
Coefficients Can Also Be Combined
Before using the quotient identity, account for coefficients on logarithms. For a positive quantity \(u\) and real constant \(c\), \(c\ln u=\ln(u^c)\). In particular, \(2\ln x=\ln(x^2)\) when \(x>0\). This can reveal a quotient whose limit is easier to study.
Worked Example: Comparing a Logarithm with Twice Another
Evaluate \(\displaystyle\lim_{x\to\infty}\bigl(\ln(x^2+1)-2\ln x\bigr)\).
For \(x>0\), rewrite \(2\ln x\) as \(\ln(x^2)\), then combine:
Both terms in the quotient tend to infinity, and both are differentiable for \(x>0\). The denominator derivative is \(2x\), which is nonzero for \(x>0\). Applying L’Hôpital’s Rule yields
Continuity of \(\ln\) at \(1\) now gives
Thus the original difference tends to \(0\). Directly, the quotient is \(1+1/x^2\), which tends to \(1\); substituting that limit into the combined logarithm confirms the answer.
A Reliable Procedure
Confirm that both arguments are positive for all sufficiently large \(x\), so the logarithms and the combining identity are valid.
Rewrite \(\ln A(x)-\ln B(x)\) as \(\ln(A(x)/B(x))\), accounting for any coefficients first.
Simplify it algebraically if possible. Use L’Hôpital’s Rule only if the quotient is still a valid indeterminate form and the rule’s conditions hold.
If the quotient tends to a positive finite number, take its logarithm by continuity. If it tends to infinity or to zero through positive values, the logarithm tends to infinity or negative infinity, respectively.
Common Mistake: Applying L’Hôpital’s Rule to a Difference
A common error is to treat \(\ln A(x)-\ln B(x)\) as if it were a quotient and take the quotient of the derivatives of its two terms. For instance, for the first example one might incorrectly write
This does not evaluate the original difference. L’Hôpital’s Rule compares derivatives only after a quotient of functions has been formed and that quotient satisfies the rule’s hypotheses. The original expression is a difference, not a quotient. The correct method is to combine the logarithms first; the resulting expression is \(\ln((x+3)/(x+1))\), whose limit is \(0\), as shown earlier.
Another potential error is to stop after differentiating the quotient inside the logarithm. Even when L’Hôpital’s Rule correctly gives the limit of that quotient, the original expression is the logarithm of the quotient. The final step must still apply the behavior of \(\ln\) to the quotient’s limit.
Why This Method Matters
Differences of diverging logarithms arise whenever two quantities grow, but their relative sizes are the real issue. Combining the logarithms makes that comparison explicit: a quotient near \(1\) produces a difference near \(0\); a quotient approaching a positive constant produces a finite difference; and a quotient that grows or shrinks without bound produces an unbounded difference. L’Hôpital’s Rule can then help compare the underlying quantities when their quotient remains indeterminate.
The central discipline is to preserve the structure of the problem. First use logarithm laws, then identify the form of the quotient, check the conditions for any differentiation-based rule, and finally translate the quotient’s limit back through the logarithm. This prevents an invalid use of L’Hôpital’s Rule and makes clear why two individually divergent expressions can have very different differences.
Check Your Understanding
For each question, state the key rewrite or condition before giving a limit.
- Why does \(\ln A(x)-\ln B(x)\) require \(A(x)>0\) and \(B(x)>0\) before it can be combined into one logarithm?
- What is the first useful rewrite for \(\ln(2x+1)-\ln(x+1)\) as \(x\to\infty\)?
- In the limit of \(\ln(x^2+1)-2\ln x\), why is it valid to replace \(2\ln x\) with \(\ln(x^2)\)?
- What conditions must be checked before applying L’Hôpital’s Rule to the quotient inside a combined logarithm?
- Why is taking the quotient of the derivatives of the two logarithmic terms not a valid way to evaluate their difference?
