Why a Difference of Divergent Expressions Needs Work
Suppose two functions both grow without bound as \(x\) increases. Their difference might also grow without bound, approach a finite number, or approach negative infinity. Knowing only that each function tends to infinity does not determine what happens to their difference.
For example, \(x-x\) is always zero, whereas \(2x-x=x\) grows without bound. These expressions share the same broad pattern: each term tends to infinity. The difference is therefore called an indeterminate form of type \(\infty-\infty\): the form alone does not tell us the limit.
L'Hôpital's Rule is a rule for certain quotients, not for differences. To use it here, first rewrite the difference as a quotient and then check that the quotient satisfies the rule's conditions. The algebraic rewrite is often the main work.
Two Useful Ways to Rewrite a Difference
One common method is to combine terms over a common denominator. Another is to multiply a difference of radicals by its conjugate. Both methods preserve the value of the expression wherever the quantities used in the rewrite are defined and the denominator is nonzero.
For a difference of squares, the identity
holds whenever \(a+b\ne 0\). Indeed, the numerator factors as \((a-b)(a+b)\), so division by \(a+b\) gives \(a-b\). For radicals, this identity is the familiar conjugate method. It can turn a difference of two large quantities into a quotient whose numerator is much simpler.
After the rewrite, the quotient must still be checked. L'Hôpital's Rule applies to a quotient in the \(0/0\) or \(\infty/\infty\) form when the numerator and denominator are differentiable near the limit point, the denominator's derivative is nonzero there, and the limit of the quotient of derivatives exists (including as an infinite limit). The hypotheses must hold on a relevant interval approaching the limit point. The rule then says the original quotient has the same limit as the derivative quotient.
Worked Example: A Radical Difference with a Finite Limit
Worked Example: A Radical Difference with a Finite Limit
Evaluate \(\displaystyle \lim_{x\to\infty}\bigl(\sqrt{x^2+3x}-x\bigr)\).
As \(x\to\infty\), both \(\sqrt{x^2+3x}\) and \(x\) tend to infinity, so the difference has the form \(\infty-\infty\). Multiply by the conjugate and simplify:
For sufficiently large positive \(x\), the denominator is positive. The transformed quotient is of type \(\infty/\infty\). Its numerator and denominator are differentiable for \(x>0\), and the derivative of the denominator is positive there, so L'Hôpital's Rule applies:
To check, divide the simplified quotient's numerator and denominator by \(x\), which is positive as \(x\to\infty\):
The check agrees with the result. The large terms in the original difference nearly cancel, leaving the finite value \(\frac{3}{2}\).
Worked Example: Combine Terms Before Applying the Rule
Worked Example: Combine Terms Before Applying the Rule
Evaluate \(\displaystyle \lim_{x\to\infty}\left(x-\frac{x^2}{x+1}\right)\).
Here, \(x\to\infty\) and \(\frac{x^2}{x+1}\to\infty\), so the difference is of type \(\infty-\infty\). Put the terms over a common denominator:
The transformed quotient is of type \(\infty/\infty\). For \(x>0\), its denominator is nonzero, and both functions are differentiable. The derivative of the denominator is \(1\), so L'Hôpital's Rule applies:
Check the result by another exact simplification:
The subtraction of two divergent quantities leaves a finite limit. Combining the terms exposed the cancellation before the limit was evaluated.
Worked Example: A Logarithmic Difference
Worked Example: A Logarithmic Difference
Evaluate \(\displaystyle \lim_{x\to\infty}\bigl(\ln(x^2+x)-2\ln x\bigr)\).
For positive \(x\), both logarithmic terms tend to infinity. Use the logarithm laws to combine the difference:
This expression tends to zero. To illustrate an application of L'Hôpital's Rule, express it as a product containing a quotient of type \(0/0\):
In the quotient, the numerator and denominator both tend to zero. For \(x>0\), both are differentiable, and the derivative of the denominator is nonzero. Differentiating the numerator and denominator gives:
Therefore, L'Hôpital's Rule gives:
The remaining factor tends to zero, so the product tends to \(1\cdot 0=0\). Thus the original difference has limit \(0\). As a check, its simplified form \(\ln(1+1/x)\) tends to \(\ln(1)=0\), since the natural logarithm is continuous at \(1\).
Why the Rewriting Works
An algebraic rewrite does not change the value of an expression on the domain where the rewrite is valid. In the radical example, multiplying by the conjugate introduced a factor equal to \(1\), provided the conjugate was nonzero. The difference of squares then reduced the numerator to a simpler expression. In the rational example, putting both terms over one denominator combined their numerators exactly.
These rewrites matter because L'Hôpital's Rule compares the rates of change of a numerator and denominator. A difference by itself is not a quotient in either of the forms covered by the rule. Once a valid quotient is obtained, the rule can apply if its hypotheses hold. The transformation does not guarantee that it will: the resulting quotient may not be \(0/0\) or \(\infty/\infty\), or another required condition may fail.
Check that each term really diverges in the direction suggested by the expression.
Try a common denominator, a conjugate, or a logarithm identity when appropriate.
Determine whether the quotient is \(0/0\) or \(\infty/\infty\), and verify the differentiability and denominator conditions.
Differentiate only after the checks, evaluate the resulting limit, and compare with a direct check when one is available.
Common Pitfall: Differentiating the Difference Directly
A frequent error is to treat L'Hôpital's Rule as permission to differentiate the terms in any indeterminate expression. For the first example, this incorrect calculation would be:
The last limit is indeed the limit of the difference of the derivatives, but L'Hôpital's Rule does not say it equals the limit of the original difference. The original expression is not a quotient of type \(0/0\) or \(\infty/\infty\). The conjugate rewrite provides an eligible quotient and gives the correct limit, \(\frac{3}{2}\).
Check Your Understanding
For each question, identify the indeterminate difference, choose a valid rewrite, and check whether L'Hôpital's Rule is applicable before using it.
- What information does the form \(\infty-\infty\) provide, and what does it leave undetermined?
- Why is multiplying \(\sqrt{x^2+5x}-x\) by its conjugate a useful first step as \(x\to\infty\)?
- Rewrite \(x-\frac{x^2}{x+2}\) as one quotient. What form does that quotient have as \(x\to\infty\)?
- What key condition makes the direct differentiation of \(\sqrt{x^2+3x}-x\) an invalid use of L'Hôpital's Rule?
- After rewriting a difference as a quotient, name two conditions you should check before applying L'Hôpital's Rule.
