Why Two Radical Terms Can Hide a Limit
A difference of two radical expressions can be hard to evaluate by direct substitution. If both radicals approach the same value, their difference approaches zero. When the denominator also approaches zero, substitution produces the form \(0/0\), which does not determine the limit. The expression may still have a finite limit: its algebra can hide a factor that cancels.
The key tools are a conjugate and, when the expression contains separate fractions, a common denominator. Used in the right order, these tools turn differences of radicals into differences of their radicands, which are often easier to simplify.
The Basic Rationalization Identity
For nonnegative \(A\) and \(B\), multiplying a difference of radicals by its conjugate gives
provided \(\sqrt{A}+\sqrt{B}\ne 0\). The identity is useful in a limit when the radicands approach positive values, so that this sum is nonzero near the limiting point. If the original difference is in a fraction, this rewrite may expose a factor that cancels with a factor in the denominator.
Cancellation is valid only where the cancelled factor is nonzero. For a limit as \(x\) approaches \(a\), it is enough that the two expressions agree for \(x\) in a deleted neighborhood of \(a\): a neighborhood around \(a\) with the point \(a\) itself excluded. Their values at \(a\) need not agree, or even be defined.
Worked Example: A Difference of Two Radicals
Worked Example: A Difference of Two Radicals
Evaluate \(\displaystyle \lim_{x\to 3}\frac{\sqrt{x+6}-\sqrt{3x}}{x-3}\). Direct substitution gives \((3-3)/(3-3)=0/0\), so first multiply the numerator by its conjugate.
For \(x\) sufficiently close to \(3\), both radicands are positive, and the sum of the radicals is nonzero. The factor \(x-3\) is nonzero in the deleted neighborhood, so its cancellation is valid there. Now the root limit and quotient law apply to the simplified expression:
Check: The rationalized expression equals the original quotient whenever \(x\ne3\) near \(3\), because multiplying the numerator by the conjugate changes its value by the factor \((\sqrt{x+6}+\sqrt{3x})/(\sqrt{x+6}+\sqrt{3x})=1\). Its limit is therefore \(-1/3\).
When the Terms Are Separate Fractions
Sometimes each radical appears in a different fraction. In that case, first combine the fractions using a common denominator. The denominators being combined must be nonzero in the nearby domain. After combining, the numerator may become a difference of radicals, which can then be rationalized.
This order matters: rationalizing before combining can make the algebra longer, while combining first may reveal a simple conjugate pair.
Worked Example: Common Denominator, Then Conjugate
Worked Example: Common Denominator, Then Conjugate
Evaluate \(\displaystyle \lim_{x\to1}\frac{\frac{1}{\sqrt{x+3}}-\frac{1}{\sqrt{3x+1}}}{x-1}\). At \(x=1\), both reciprocal terms are \(1/2\), so the numerator and denominator both approach zero.
For \(x\) near \(1\), the two radicands are positive. Combine the fractions using the common denominator \(\sqrt{x+3}\sqrt{3x+1}\):
Now multiply the numerator and denominator by the conjugate of the numerator, \(\sqrt{3x+1}+\sqrt{x+3}\):
The original fraction denominators are nonzero near \(1\), and the new common denominator is positive there. Cancelling \(x-1\) is valid for \(x\ne1\). The simplified expression has a nonzero denominator at \(1\), so substitution gives
Check: The combined numerator before rationalization is \(\sqrt{3x+1}-\sqrt{x+3}\). Multiplying it by its conjugate gives \((3x+1)-(x+3)=2(x-1)\), exactly the factor cancelled against the original denominator. Thus the simplified expression has limit \(1/8\).
Rationalizing Both the Numerator and Denominator
A zero-over-zero quotient can have a difference of radicals in both its numerator and its denominator. In that case, rationalize each difference separately. The resulting expressions may have factors that cancel, leaving a quotient of sums of radicals. Check that each conjugate sum is nonzero near the point before using the rewrite.
Worked Example: Two Conjugate Rewrites
Worked Example: Two Conjugate Rewrites
Evaluate \(\displaystyle \lim_{x\to1}\frac{\sqrt{2x+2}-\sqrt{3x+1}}{\sqrt{x+8}-3}\). At \(x=1\), the two numerator radicals are both \(2\), and the denominator is \(3-3\), so direct substitution gives \(0/0\).
Rationalize the numerator difference and the denominator difference. The conjugate of the denominator's difference is \(\sqrt{x+8}+3\):
Now multiply by \((\sqrt{x+8}+3)/(\sqrt{x+8}+3)\). Since \((\sqrt{x+8}-3)(\sqrt{x+8}+3)=x-1\), and \(1-x=-(x-1)\), the factors simplify:
Near \(1\), all radicands are positive, and both conjugate sums used in the rationalizations are nonzero. The cancelled factor \(x-1\) is nonzero for \(x\ne1\). Substitute \(x=1\) into the simplified expression:
Check: Rationalizing the numerator produces \(1-x=-(x-1)\); rationalizing the denominator produces \(x-1\). Their quotient contributes the factor \(-1\), so the final negative sign is required. The simplified expression tends to \(-3/2\), as claimed.
Why the Method Works
The method rests on an exact algebraic identity, not on an approximation. For \(A,B\ge0\) and \(\sqrt A+\sqrt B\ne0\),
Thus, if the radicands are functions of \(x\), rationalizing changes the radical difference into the difference of those functions. When both radicands have the same value at \(x=a\), their difference is zero at \(a\); after factoring, it may contain \(x-a\). That factor can cancel the denominator's \(x-a\), leaving an expression whose denominator is nonzero at \(a\).
Here is the complete algebraic justification for the central step. Suppose \(A(x)\) and \(B(x)\) are nonnegative near \(a\), their radical sum is nonzero near \(a\), and \(x\ne a\). Then
If \(A(x)-B(x)=(x-a)C(x)\), this becomes
If \(C(x)\) and the radical sum have limits at \(a\), and the radical sum's limit is nonzero, the quotient law evaluates the limit of the right-hand side. The assumptions matter: the radicals must be defined near \(a\), and the denominator of the simplified quotient must have a nonzero limit for this direct use of the quotient law.
Common Mistake: Substituting into \(0/0\)
A common error is to treat the result \(0/0\) as though it were the value of the limit, or to conclude that the limit does not exist. It is neither a numerical value nor, by itself, evidence that a limit fails to exist. It signals that direct substitution has not finished the calculation.
For instance, in the first example, substituting \(x=3\) directly gives
Calling the limit “undefined” at this stage is incorrect. The expression is undefined at \(x=3\), but the limit concerns values of \(x\) near \(3\), not the value at \(3\). Rationalization shows that for \(x\ne3\) the expression equals \(-2/(\sqrt{x+6}+\sqrt{3x})\), which tends to \(-1/3\).
A Reliable Order of Operations
Identify the limiting point, see whether the radicals are defined nearby, and note whether substitution gives \(0/0\).
Use a common denominator when the expression contains terms in separate fractions, checking that the denominators are nonzero near the point.
Choose the conjugate of the radical difference that prevents direct substitution from resolving the limit.
Show the factorization and state that the cancelled factor is nonzero for nearby \(x\ne a\).
Apply limit laws only after the simplified denominator has a nonzero limit, then verify the key algebraic identity.
Check Your Understanding
For each problem, identify the useful conjugate or common denominator before evaluating the limit.
- Evaluate \(\displaystyle \lim_{x\to2}\frac{\sqrt{x+7}-\sqrt{3x+3}}{x-2}\). What factor appears after rationalizing?
- In \(\displaystyle \lim_{x\to1}\frac{\frac1{\sqrt{x+8}}-\frac1{\sqrt{2x+7}}}{x-1}\), why is it useful to find a common denominator first?
- State the condition that permits cancellation of a factor \(x-a\) when evaluating a limit as \(x\to a\).
- Why does obtaining \(0/0\) by substitution not determine whether the limit exists?
- For a difference \(\sqrt{A(x)}-\sqrt{B(x)}\), what expression results in the numerator after multiplying by its conjugate?
