Why Conjugates Help with Radical Differences
A difference such as \(\sqrt{x^2+6x+5}-\sqrt{x^2+2x-3}\) has the form infinity minus infinity as \(x\to\infty\). Each root grows without bound, but their difference can approach a finite number. Estimating each term separately does not reveal that number: the important information is the small difference between the radicands.
Multiplying by a conjugate exposes that information. The conjugate of \(\sqrt{A(x)}-\sqrt{B(x)}\) is \(\sqrt{A(x)}+\sqrt{B(x)}\). Their product is the difference of the radicands, because \((u-v)(u+v)=u^2-v^2\). This converts the original difference into a quotient that is often much easier to evaluate.
The Conjugate Identity
To verify the identity, multiply the original difference by a form of \(1\), namely the conjugate divided by itself. The numerator becomes a difference of squares:
This is an algebraic identity, not an approximation. It applies at every point where its square roots and denominator are defined. For a limit at infinity, it is enough that these conditions hold for all sufficiently large values of \(x\) (or sufficiently negative values, for a limit at negative infinity).
The key is to compare the growth of the new numerator and denominator. If \(A(x)-B(x)\) grows roughly like \(x\), while each root grows roughly like \(x\) as \(x\to\infty\), the quotient may approach a finite value. If the radicand difference grows faster than the sum of the roots, the radical difference may instead grow without bound.
Worked Example: A Finite Limit from Two Growing Roots
Worked Example: A Finite Limit from Two Growing Roots
Evaluate \[ \lim_{x\to\infty}\left(\sqrt{x^2+6x+5}-\sqrt{x^2+2x-3}\right). \]
For sufficiently large \(x\), both radicands are positive, and both square roots tend to infinity. Multiply by the conjugate:
Since \(x\to\infty\), divide the numerator and denominator by \(x\), which is positive in this limit:
Each fraction involving \(x\) in the roots tends to zero. By continuity of the square-root function at \(1\), each root in the denominator tends to \(1\). Thus the numerator tends to \(4\), and the denominator tends to \(1+1=2\). The limit is \(4/2=2\).
Check: the conjugate identity is exact, and substituting the limiting values into the transformed quotient gives \(4/2=2\). The original expression therefore has the same limit.
How to Choose Between Conjugates and L'Hôpital's Rule
After rationalizing, the expression has the form \[ \frac{A(x)-B(x)}{\sqrt{A(x)}+\sqrt{B(x)}}. \] This is the algebraic conversion that makes the cancellation visible. Sometimes its limit can be evaluated directly by dividing through by the dominant power of \(x\), as in the preceding example.
The transformed expression can also be a quotient of type infinity over infinity. In that case, L'Hôpital's Rule may be available, but it is not automatic: the numerator and denominator must be differentiable on an appropriate interval, the denominator derivative must be nonzero there, and the quotient of derivatives must have a limit. The quotient's derivative is not found by differentiating it as a whole; L'Hôpital's Rule uses the ratio of the numerator derivative to the denominator derivative.
A practical distinction is that conjugate multiplication is an identity tailored to square-root differences. It is often shorter and makes the scale of the answer clear. L'Hôpital's Rule is an additional option only after the transformed expression satisfies its hypotheses. If the transformed quotient is not an indeterminate form, or its derivative ratio is less manageable, use direct algebra instead.
Worked Example: Conjugates and L'Hôpital Give the Same Limit
Worked Example: Conjugates and L'Hôpital Give the Same Limit
Evaluate \[ \lim_{x\to\infty}\left(\sqrt{9x^2+4x+2}-3x\right). \]
For sufficiently large \(x\), the radicand is positive. Multiplying by the conjugate gives
In the last line, \(x>0\), so factoring \(x^2\) from under the root gives \(\sqrt{x^2(9+4/x+2/x^2)}=x\sqrt{9+4/x+2/x^2}\). The numerator tends to \(4\), and the denominator tends to \(\sqrt{9}+3=6\). Therefore the limit is \(4/6=2/3\).
For comparison, apply L'Hôpital's Rule to the conjugate quotient \((4x+2)/(\sqrt{9x^2+4x+2}+3x)\). Its numerator and denominator both tend to infinity. The denominator is differentiable and its derivative is positive for sufficiently large \(x\). Differentiating the numerator and denominator separately gives
Here \(\sqrt{9x^2+4x+2}/x\to 3\), so \((18x+4)/(2\sqrt{9x^2+4x+2})\to 9/3=3\). The result agrees with the direct conjugate calculation.
When the Difference Does Not Approach a Finite Number
Conjugate multiplication does not guarantee a finite limit. It reveals the relative growth of the radicand difference and the sum of the roots, which may show that the original difference tends to positive or negative infinity.
Worked Example: A Radical Difference That Grows Without Bound
Evaluate \[ \lim_{x\to\infty}\left(\sqrt{4x^2+x+1}-\sqrt{x^2+2x+3}\right). \]
Both radicands are positive for sufficiently large \(x\), so conjugate multiplication gives
For \(x>0\), divide the numerator by \(x^2\) and the denominator by \(x\). The quotient can then be written as
The fraction tends to \(3/(2+1)=1\), while \(x\to\infty\). The entire expression therefore grows like \(x\) and tends to positive infinity. More explicitly, the fraction is positive and tends to \(1\), so it is greater than \(1/2\) for all sufficiently large \(x\); the product is then greater than \(x/2\), which tends to infinity. The limit is \(+\infty\).
Check: the leading terms of the radicands differ by \(3x^2\), while the sum of the roots grows like \(3x\). Their quotient consequently grows like \(x\), consistent with the computed divergence.
A Sign Pitfall at Negative Infinity
When \(x\to-\infty\), it is essential to remember that \(\sqrt{x^2}=|x|=-x\), not \(x\). Factoring \(x^2\) from a square root gives \(\sqrt{x^2C(x)}=|x|\sqrt{C(x)}\) whenever \(C(x)\ge 0\). Using \(x\sqrt{C(x)}\) at negative \(x\) introduces a sign error.
For example, consider \(\sqrt{x^2+3x}-\sqrt{x^2-x}\) as \(x\to-\infty\). A wrong calculation might factor out \(x\) from both roots, claim their sum behaves like \(2x\), and conclude that the conjugate quotient tends to \(2\). The correction starts with the exact identity:
For sufficiently negative \(x\), factor out \(-x=|x|\) from each root. Dividing the quotient by this positive scale gives
Thus the correct limit is \(-2\). The sign comes from the numerator \(4x\), while the denominator is positive. Factoring out \(|x|\) preserves both facts.
A Reliable Calculation Sequence
Confirm both radicands are nonnegative and that the conjugate denominator is nonzero near the limiting point.
Replace the difference of roots by the difference of radicands divided by the sum of the roots.
For limits at infinity, factor out the appropriate power of \(x\), using \(|x|\) when taking a square root of \(x^2\).
Evaluate directly if the simplified quotient has an accessible limit. Use L'Hôpital's Rule only if its form and differentiability conditions hold.
Why This Method Matters
The main difficulty in many radical infinity differences is cancellation: each term can be large, while their difference is small. Conjugate multiplication removes the subtraction between the roots and replaces it with a subtraction between their radicands. That subtraction may be a simple polynomial expression, making the limit visible.
The method also helps diagnose the answer before completing every algebraic step. If the radicands have the same leading term, their difference may grow more slowly than either radicand, allowing the radical difference to have a finite limit. If their leading terms differ, the difference of the roots may instead grow without bound. These observations guide the calculation, but the identity and a valid limit argument establish the result.
A common mistake is to assume that any infinity-minus-infinity expression should be handled with L'Hôpital's Rule. The conjugate identity is exact and often resolves the radical structure directly. Another mistake is to apply L'Hôpital's Rule to the original difference without first converting it into a quotient: the rule concerns particular quotient forms, not a difference of two functions that each diverge.
Check Your Understanding
Use the conjugate identity where appropriate, and state any domain or sign condition that matters.
- Evaluate \(\lim_{x\to\infty}\left(\sqrt{x^2+8x+1}-\sqrt{x^2+2x+4}\right)\).
- For \(\sqrt{A(x)}-\sqrt{B(x)}\), what conditions make the conjugate identity valid?
- Why does \(\sqrt{x^2}=|x|\) matter when evaluating a limit as \(x\to-\infty\)?
- What must be checked before applying L'Hôpital's Rule to the quotient obtained after conjugate multiplication?
- Does conjugate multiplication guarantee a finite limit? Explain what the transformed quotient reveals.
