Why Rationalize a Radical Numerator?
A quotient can have a radical numerator that approaches zero at the same input as its denominator. Direct substitution then gives the indeterminate form \(0/0\), which does not determine the limit. Multiplying the numerator by its conjugate can transform the radical difference into a difference of squares. That algebra may reveal a factor that cancels with a factor in the denominator.
The method is especially useful when the expression has the form \(\sqrt{g(x)}-c\), where \(g(x)\) approaches \(c^2\). The conjugate is \(\sqrt{g(x)}+c\). The key identity is:
This tutorial focuses on rationalizing the numerator. The goal is not to remove every radical from an expression, but to expose a factor that makes the limit easier to evaluate.
The Algebraic Step and Its Conditions
Multiplying a quotient by a factor is valid only if the factor equals \(1\). Thus, when the conjugate is defined and nonzero, rationalizing the numerator means multiplying both the numerator and denominator by the conjugate:
The first equality holds wherever the original quotient is defined and \(\sqrt{g(x)}+c\ne 0\). In the common case \(c>0\), the conjugate is positive wherever the square root is defined, so it cannot be zero.
For the limit method, suppose \(g\) is continuous at \(a\), \(g(a)=c^2\), and \(c>0\). Also suppose \(g(x)\ge 0\) for \(x\) in a neighborhood of \(a\), so the square root is defined there. Continuity gives \(g(x)\to c^2\), and continuity of the square-root function on nonnegative inputs gives:
This conjugate factor therefore approaches a nonzero number. Continuity matters: knowing only the value \(g(a)=c^2\) does not by itself determine what \(g(x)\) approaches as \(x\to a\).
When the Cancellation Produces a Limit
A common pattern is that both \(g(x)-c^2\) and \(h(x)\) contain a factor \(x-a\). For example, suppose near \(a\) they can be written as
where \(q\) and \(r\) are continuous at \(a\), and \(r(a)\ne 0\). For \(x\ne a\) sufficiently close to \(a\), rationalizing and cancelling gives:
The cancelled expression agrees with the original quotient for nearby \(x\ne a\). By the limit laws and the continuity assumptions, its limit is:
The cancellation is not a claim that the original quotient is defined at \(x=a\). It simplifies the expression for nearby inputs different from \(a\), which is enough to determine the limit.
Worked Examples
Worked Example: A Square Root of a Linear Expression
Evaluate \(\displaystyle \lim_{x\to 4}\frac{\sqrt{x+5}-3}{x-4}\).
Direct substitution gives \(\sqrt{4+5}-3=3-3=0\) in the numerator and \(4-4=0\) in the denominator. Rationalize the numerator by multiplying both parts of the quotient by its conjugate, \(\sqrt{x+5}+3\):
Apply the difference-of-squares identity in the numerator:
For \(x\ne 4\), cancel the common factor \(x-4\):
Now use continuity of the square-root function and substitute \(x=4\) into the simplified expression:
Check: For \(x\ne4\), the original quotient equals \(1/(\sqrt{x+5}+3)\). As \(x\to4\), that expression approaches \(1/6\), confirming the calculated limit. The conjugate approaches \(6\), not zero.
Worked Example: A Linear Expression Inside the Radical
Evaluate \(\displaystyle \lim_{x\to 4}\frac{\sqrt{2x+1}-3}{x-4}\).
Substitution gives \(\sqrt{2(4)+1}-3=\sqrt{9}-3=0\) and \(4-4=0\). Multiply numerator and denominator by \(\sqrt{2x+1}+3\):
Factor the numerator completely:
For \(x\ne4\), cancel \(x-4\), then evaluate the remaining limit:
Check: The rationalized expression equals the original one for \(x\ne4\). Its denominator approaches \(6\), so its value approaches \(2/6=1/3\), as found.
Worked Example: Cancelling a Factor from a Quadratic Denominator
Evaluate \(\displaystyle \lim_{x\to-3}\frac{\sqrt{x+7}-2}{x^2-9}\).
At \(x=-3\), the numerator is \(\sqrt{4}-2=0\), and the denominator is \(9-9=0\). Multiply the numerator and denominator by the conjugate \(\sqrt{x+7}+2\):
Factor the difference of squares in the denominator:
For \(x\ne-3\), cancel \(x+3\):
The remaining expression is defined and continuous at \(-3\). Substitution gives:
Check: Near \(x=-3\), the rationalized numerator is \(x+3\), while the denominator has the factor \(x+3\). After cancellation, the other denominator factors approach \(-6\) and \(4\), whose product is \(-24\). The limit is therefore \(-1/24\).
Why the Conjugate Works
The method works because the conjugate converts a difference of square roots into an ordinary algebraic difference. The identity can be verified by multiplication:
Setting \(u=\sqrt{g(x)}\) and \(v=c\) gives \(g(x)-c^2\). When \(g\) is a polynomial or another expression whose difference \(g(x)-g(a)\) contains \(x-a\), this transformation can expose the factor needed to cancel a zero in the denominator.
The limit step then follows from the limit laws: after cancellation, if the remaining denominator approaches a nonzero number, direct substitution into the simplified expression evaluates the limit. This is not direct substitution into the original \(0/0\) form; it is substitution after an algebraically justified simplification.
A Common Mistake: Forgetting the Denominator
A frequent error is to multiply only the numerator by the conjugate and then use the difference-of-squares identity as if the quotient were unchanged. For the first example, an incorrect calculation might say:
This is wrong because the numerator was changed without making the same change to the denominator. The fraction was not multiplied by \(1\). The correct step multiplies both parts by the conjugate:
Consequently, the limit is \(1/6\), not \(1\). Always check that the transformation preserves the whole quotient before cancelling.
Using the Method Carefully
Before rationalizing, identify the exact radical binomial in the numerator and write its conjugate by changing the sign between its two terms. After multiplying both parts of the quotient, simplify the product using the difference-of-squares identity. Then factor the resulting numerator and denominator and cancel only a factor that is nonzero for the nearby inputs under consideration.
Confirm that direct substitution produces a form such as \(0/0\), with a radical difference in the numerator.
Multiply both numerator and denominator by the same conjugate, so the quotient is unchanged wherever the multiplier is defined and nonzero.
Use the difference of squares, factor any common term, and cancel it for nearby inputs other than the limiting input.
Use the limit laws on the simplified expression, checking that its denominator approaches a nonzero value.
Do not assume cancellation will always produce a finite limit. If no common factor appears, if the remaining denominator still approaches zero, or if a required square root is not defined near the input, more analysis is needed. Also, when the constant \(c\) is zero, the conjugate \(\sqrt{g(x)}+c\) may approach zero; the nonzero-conjugate argument given here then does not apply automatically.
Check Your Understanding
For each limit, decide how the numerator's conjugate changes the expression and show the cancellation, if any.
- Evaluate \(\displaystyle \lim_{x\to 3}\frac{\sqrt{x+6}-3}{x-3}\).
- Evaluate \(\displaystyle \lim_{x\to 1}\frac{\sqrt{3x+1}-2}{x-1}\).
- For a numerator \(\sqrt{g(x)}-c\), state why continuity of \(g\), \(g(a)=c^2\), and \(c>0\) imply that the conjugate approaches a nonzero value.
- Explain why cancelling a factor \(x-a\) can help find a limit even when the original quotient is undefined at \(x=a\).
- Identify the error in multiplying only the numerator of a quotient by its conjugate.
