Why Infinity Minus Infinity Needs Rewriting
When two quantities both grow without bound, their difference may approach a finite number, grow without bound, or fail to have a limit. Knowing only that each term tends to infinity does not tell us what happens to their difference.
For example, \(x\) and \(\frac{x^2}{x+1}\) both tend to infinity as \(x\to\infty\). Yet their difference has a finite limit. The large terms nearly cancel, so the important information is in what remains after cancellation.
This is the indeterminate form called infinity minus infinity, written \(\infty-\infty\). It is not a value that can be calculated by subtracting infinities. Instead, it signals that the expression must be rewritten before its limit can be determined.
Combining Terms Over a Common Denominator
For rational expressions, a common denominator often reveals the cancellation directly. If \(V(x)\) and \(Z(x)\) are nonzero near the endpoint, then
The numerator must be expanded or factored carefully. In particular, the minus sign applies to the entire second product. After combining, the new numerator and denominator may have limits that are no longer indeterminate. In that case, use ordinary limit laws rather than L’Hôpital’s Rule.
Worked Example: A Finite Limit Hidden by Two Diverging Terms
Evaluate \(\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x\right)\). Both terms tend to \(+\infty\), so the original difference has the form \(\infty-\infty\).
Write \(x\) with denominator \(x+1\), which is nonzero for \(x\to\infty\):
Expand the numerator, keeping the subtraction sign:
Thus the expression is \(-\frac{x}{x+1}\). The quotient \(\frac{x}{x+1}\) has the form \(+\infty/+\infty\). For \(x>0\), its numerator and denominator are differentiable, and the derivative of the denominator is \(1\), which is nonzero. L’Hôpital’s Rule therefore gives
Consequently, the original limit is \(-1\). Check by dividing numerator and denominator of the simplified expression by \(x\):
A Quotient Conversion for Differences
There is also an algebraic conversion that works when one of the terms is nonzero. If \(g(x)\ne 0\), then
To verify the identity, combine the numerator on the right:
If \(f(x)\to+\infty\) and \(g(x)\to+\infty\), then \(1/g(x)\to 0\). The numerator \(f(x)/g(x)-1\) may tend to zero, a nonzero number, or infinity. Therefore, the conversion alone does not guarantee an indeterminate quotient. Check the limits of the new numerator and denominator before applying L’Hôpital’s Rule.
In particular, if \(f/g\to 1\), the new quotient has the form \(0/0\). L’Hôpital’s Rule may then be useful, provided the functions are differentiable near the endpoint, the denominator’s derivative is nonzero there, and the derivative quotient has a limit. These conditions are essential; algebraic rewriting does not by itself establish them.
Worked Example: Using the Quotient Conversion to Find a Finite Limit
Evaluate \(\displaystyle\lim_{x\to\infty}\bigl((x^2+2x+3)-(x^2+2x)\bigr)\). Each parenthesized quantity tends to \(+\infty\), so the original form is \(\infty-\infty\). Set \(f(x)=x^2+2x+3\) and \(g(x)=x^2+2x\). For \(x>0\), \(g(x)>0\), so the conversion is valid:
Combine the numerator over its denominator:
The converted expression is therefore
Here both numerator and denominator of the outer quotient tend to \(0\), so its form is \(0/0\). We can apply L’Hôpital’s Rule: \(x^2+2x>0\) and its derivative \(2x+2\) is nonzero for \(x>0\). Differentiating the outer numerator and denominator gives
The limit is \(3\). The result agrees with direct subtraction: \((x^2+2x+3)-(x^2+2x)=3\).
What L’Hôpital’s Rule Requires
For the cases used here, L’Hôpital’s Rule says the following. Suppose \(F\) and \(G\) are differentiable on an interval extending toward the endpoint, \(G'(x)\ne 0\) sufficiently near that endpoint, and both \(F(x)\) and \(G(x)\) tend to \(0\), or both tend to \(+\infty\). If \(\frac{F'(x)}{G'(x)}\) has a limit \(L\), finite or infinite, then \(\frac{F(x)}{G(x)}\) has the same limit \(L\).
This rule concerns a quotient in an appropriate indeterminate form. It does not say that differentiating the two terms of a difference preserves that difference’s limit. When a conversion produces a quotient, verify its form and check that the derivative quotient has the required limit before drawing a conclusion.
Worked Example: A Difference That Still Diverges
Evaluate \(\displaystyle\lim_{x\to\infty}\bigl((x^2+x)-x^2\bigr)\). Both terms tend to \(+\infty\). Use \(f(x)=x^2+x\) and \(g(x)=x^2\), with \(g(x)>0\) for \(x>0\):
Simplify the numerator:
The resulting quotient is \(\frac{1/x}{1/x^2}\). Its numerator and denominator both tend to \(0\), so it has the form \(0/0\). For \(x>0\), the denominator function \(1/x^2\) has derivative \(-2/x^3\ne 0\). Apply L’Hôpital’s Rule:
Thus the original difference tends to \(+\infty\), even though both of its terms tend to infinity. Check the result by direct subtraction:
A Common Mistake: Differentiating the Difference
A tempting but invalid move is to differentiate each term of an \(\infty-\infty\) expression and treat the result as the original limit. For the first worked example, this would give
This would incorrectly suggest that the original limit is \(0\), although its value is \(-1\). L’Hôpital’s Rule does not permit differentiating the terms of a difference in this way. Its numerator and denominator must be parts of a quotient in the required indeterminate form.
The correction is to combine the original terms first:
Only after obtaining a quotient may you check whether L’Hôpital’s Rule applies to it. Even then, a direct algebraic simplification or ordinary limit law may be simpler.
A Reliable Procedure
Check that each term tends to infinity. This identifies \(\infty-\infty\) as indeterminate; it does not determine the difference’s limit.
For rational expressions, combine over a common denominator. Alternatively, use \((f/g-1)/(1/g)\) when \(g\ne 0\).
Simplify and find the separate limits of the quotient’s numerator and denominator. Do not assume the result is \(0/0\) or \(\infty/\infty\).
Check the indeterminate form, differentiability, the nonvanishing denominator derivative near the endpoint, and the limit of the derivative quotient.
Check signs and cancellation in the algebra, and substitute the resulting limit into the simplified expression or compare with a direct simplification.
Why the Rewriting Matters
The original terms in an infinity-minus-infinity expression can be individually large while their difference is small, constant, or still unbounded. Rewriting exposes the cancellation that determines the result. A common denominator is especially effective for rational expressions; the quotient conversion can be useful when the two terms have a ratio near \(1\).
The central discipline is to separate two tasks: first, rewrite the difference correctly; then, decide how to evaluate the resulting expression. L’Hôpital’s Rule is one possible tool, not an automatic response to the phrase “infinity minus infinity.”
Check Your Understanding
For each question, identify the transformed form before deciding whether L’Hôpital’s Rule applies.
- Why does knowing that \(f(x)\to+\infty\) and \(g(x)\to+\infty\) not determine the limit of \(f(x)-g(x)\)?
- Combine \(\displaystyle\frac{x^2+1}{x+1}-x\) over a common denominator and simplify.
- State the identity that rewrites \(f-g\) as a quotient involving \(f/g-1\), and give the condition needed for it to be valid.
- For \(\displaystyle\lim_{x\to\infty}\bigl((x^2+2x+3)-(x^2+2x)\bigr)\), what form does the quotient conversion produce?
- Why is differentiating the two terms of an infinity-minus-infinity difference not an application of L’Hôpital’s Rule?
