Why Rationalize a Denominator?
A quotient may be undefined at the point of interest because its denominator approaches zero. If the numerator approaches zero as well, direct substitution gives the indeterminate form \(0/0\), so it does not determine the limit. When the denominator contains a difference of square roots, multiplying by its conjugate can turn that denominator into a difference of radicands. This may reveal a factor that can be canceled, leaving an expression whose limit can be evaluated directly.
The key is to make an equivalent algebraic rewrite for nearby values of \(x\), not to assign a value to the original quotient at the point where it is undefined. The rewrite is useful only when its domain and cancellation conditions are respected.
Conjugates and the Difference-of-Squares Identity
The conjugate of an expression \(A-B\) is \(A+B\); conversely, the conjugate of \(A+B\) is \(A-B\). The reason conjugates help is the difference-of-squares identity:
For square roots, squaring removes the radicals: \((\sqrt{u})^2=u\) when \(u\geq 0\). Thus the conjugate of \(\sqrt{u(x)}-\sqrt{v(x)}\) is \(\sqrt{u(x)}+\sqrt{v(x)}\), and their product is \(u(x)-v(x)\).
This multiplication preserves the value of the quotient only where the conjugate is defined and nonzero. For a limit at \(x=a\), the rewrite must hold for \(x\) in a deleted neighborhood of \(a\): nearby values other than \(a\), where the original quotient is defined. If cancellation follows, the canceled expression agrees with the original quotient at those nearby points. It need not agree at \(x=a\), where the original quotient may be undefined.
When a Radical Denominator Approaches Zero
A common denominator has the form \(\sqrt{g(x)}-c\), where \(c\) is a constant. If \(g\) is continuous at \(a\), \(g(x)\geq 0\) near \(a\), and \(\sqrt{g(x)}-c\to 0\) as \(x\to a\), then continuity of the square-root function gives \(\sqrt{g(a)}-c=0\). In this situation, \(c\geq 0\) and \(g(a)=c^2\). Continuity is important: a limit of \(\sqrt{g(x)}-c\) alone does not establish what \(g(a)\) equals if \(g\) is discontinuous at \(a\).
When \(c>0\), the conjugate \(\sqrt{g(x)}+c\) is positive wherever the square root is defined, so multiplication by it is valid there. If the conjugate is zero, or if a radicand is negative, the proposed rewrite may not be defined; check these conditions rather than applying the rule automatically.
A General Justification
Consider a quotient with denominator \(\sqrt{u(x)}-\sqrt{v(x)}\). Suppose both radicands are nonnegative near \(a\), the original quotient is defined for nearby \(x\ne a\), and the conjugate \(\sqrt{u(x)}+\sqrt{v(x)}\) is nonzero at those nearby values. Multiplying top and bottom by that conjugate gives:
The last equality follows from the difference-of-squares identity. This is an equality wherever both quotients are defined, not a claim that either quotient has a value at \(x=a\). If \(u(x)-v(x)\) has a factor shared with \(N(x)\), that factor can be canceled for nearby \(x\ne a\). The simplified expression can then be used to evaluate the limit, provided it has a limit there. This argument explains why the method works: it replaces the original quotient by an equal quotient at nearby points whose denominator may be easier to analyze.
Worked Examples
Worked Example: A Linear Numerator and a Single Radical
Evaluate \(\displaystyle \lim_{x\to 4}\frac{x-4}{\sqrt{x+5}-3}\).
First substitute \(x=4\) into the numerator and denominator. The numerator is \(4-4=0\), and the denominator is \(\sqrt{4+5}-3=3-3=0\). The quotient has the form \(0/0\), so direct substitution does not evaluate the limit.
The denominator’s conjugate is \(\sqrt{x+5}+3\). Multiply both numerator and denominator by it:
The cancellation is valid for \(x\ne4\), which is exactly the nearby setting used for the limit. Now evaluate the limit of the simplified expression:
Check: For \(x\ne4\) near \(4\), the original quotient equals \(\sqrt{x+5}+3\). As \(x\to4\), this expression approaches \(6\), so the limit is \(6\); no value at \(x=4\) was assigned to the original quotient.
Worked Example: A Radical Denominator with a Variable Coefficient
Evaluate \(\displaystyle \lim_{x\to0}\frac{x}{\sqrt{1+3x}-1}\).
At \(x=0\), the numerator is \(0\), and the denominator is \(\sqrt{1}-1=0\). The radicand \(1+3x\) is positive for \(x\) sufficiently close to \(0\), so the square root is defined on a neighborhood of the limit point.
The conjugate of \(\sqrt{1+3x}-1\) is \(\sqrt{1+3x}+1\). Multiplying both parts of the quotient gives:
Now substitute \(x=0\) into the simplified expression:
Check: For nearby \(x\ne0\), the denominator in the original quotient is nonzero and the algebra above gives the same quotient as \((\sqrt{1+3x}+1)/3\). That expression approaches \(2/3\), verifying the limit.
Worked Example: Two Variable Radicals in the Denominator
Evaluate \(\displaystyle \lim_{x\to2}\frac{x-2}{\sqrt{x+3}-\sqrt{2x+1}}\).
At \(x=2\), the numerator is \(2-2=0\). The radicands in the denominator are \(2+3=5\) and \(2(2)+1=5\), so the denominator is \(\sqrt5-\sqrt5=0\). Both radicands are positive near \(2\).
The conjugate of the denominator is \(\sqrt{x+3}+\sqrt{2x+1}\). Multiply by it in both numerator and denominator:
In the last step, \(2-x=-(x-2)\), so \((x-2)/(2-x)=-1\) for \(x\ne2\). Evaluate the limit of the resulting expression:
Check: The rationalized quotient equals the original quotient for nearby \(x\ne2\). Its limit is \(-2\sqrt5\), including the negative sign produced by \(2-x=-(x-2)\).
A Common Mistake: Multiplying Only the Denominator
In the second example, multiplying the denominator \(\sqrt{1+3x}-1\) by its conjugate produces \(3x\). It is incorrect to change the original quotient into \(x/(3x)\) and conclude that the limit is \(1/3\). That step changes only the denominator, so it changes the quotient.
The valid operation multiplies both numerator and denominator by the same conjugate:
The conjugate factor in the numerator is essential. Omitting it loses the factor that makes the limit \(2/3\), rather than the incorrect \(1/3\).
How to Apply the Method Carefully
Check the numerator and denominator at the limit point. If both approach zero, recognize that substitution alone does not determine the limit.
Change the sign between the two terms, keeping the terms themselves unchanged.
Use the same conjugate in numerator and denominator, and check that it is defined and nonzero for the nearby values used.
Apply the difference-of-squares identity. Cancel a shared factor only for values where that factor is nonzero.
Find the limit of the simplified expression, then confirm that it agrees with the original quotient for nearby points in its domain.
Why the Conditions Matter
Rationalizing is an algebraic rewrite, not a rule that guarantees every limit exists or is finite. The resulting expression may still have a zero denominator, or the conjugate used as a multiplier may fail to be defined or nonzero. In those cases, this particular rewrite has not yet made the limit evaluable.
Also distinguish a function’s limit from its value at the limit point. A quotient that has the form \(0/0\) at \(x=a\) is undefined there, even if rationalization produces a simpler expression that has a value at \(a\). The limit concerns the values at nearby points, not the original quotient’s value at \(a\). This distinction is why each example states that cancellation is for \(x\ne a\).
Check Your Understanding
For each problem, first check the form obtained by direct substitution. Then decide whether multiplying by a denominator’s conjugate helps, and keep track of where any cancellation is valid.
- Evaluate \(\displaystyle \lim_{x\to3}\frac{x-3}{\sqrt{x+6}-3}\).
- What is the conjugate of \(\sqrt{4x+1}-\sqrt{x+7}\)?
- In a quotient with denominator \(\sqrt{h(x)}-2\), what additional condition lets you infer \(h(a)=4\) from \(\sqrt{h(x)}-2\to0\) as \(x\to a\)?
- Why must the numerator and denominator both be multiplied by the same conjugate?
- When may a shared factor be canceled in evaluating a limit at \(x=a\), even if the factor is zero at \(a\)?
