Why Multiply by a Conjugate?
A difference of square roots can hide a factor that is difficult to see by direct substitution. For example, when \(x\) approaches \(4\), both \(\sqrt{x+5}-3\) and \(x-4\) approach \(0\). Substituting \(x=4\) into their quotient therefore gives the indeterminate form \(0/0\), not the value of the limit.
The useful algebraic move is to multiply by the expression with the same two terms but the opposite sign between them. This is called the conjugate. The product of a difference and its conjugate is a difference of squares, which can reveal a factor matching the denominator.
The Rationalization Method
Consider a limit of the form \[ \lim_{x\to a}\frac{\sqrt{u(x)}-\sqrt{v(x)}}{x-a}, \] where \(u(x)\) and \(v(x)\) are defined and nonnegative near \(a\). If direct substitution gives \(0/0\), then \(u(a)=v(a)\). Multiply the numerator and denominator by the conjugate of the numerator:
This multiplication does not change the expression wherever the added denominator is nonzero. If the radicands are nonnegative and at least one is positive, their square roots have a positive sum. In the common case where both approach the same positive number, the sum is certainly nonzero near \(a\).
The next step is to factor \(u(x)-v(x)\). When that difference contains \(x-a\), cancel the shared factor, then evaluate the limit of the resulting expression. Cancellation is valid for \(x\ne a\); a limit depends on values near \(a\), not on whether the original expression is defined at \(a\) itself.
Why the Method Works
The justification comes directly from the difference-of-squares identity. If \(A=\sqrt{u(x)}\) and \(B=\sqrt{v(x)}\), then multiplying by \(A+B\) replaces \(A-B\) with \(A^2-B^2=u(x)-v(x)\). This removes the radicals from the difference. If the radicand difference has a factor \(x-a\), the algebra exposes the factor needed to simplify the quotient.
Confirm that the expression has the form \(0/0\) before using rationalization to resolve it.
Use the conjugate of the difference of square roots, multiplying both numerator and denominator by it.
Replace the product of the conjugates with the difference of the radicands.
Factor the radicand difference and cancel \(x-a\) only for \(x\ne a\).
Use direct substitution if the simplified expression is continuous and defined at \(a\).
For a common special case, the radicands are linear functions and agree at the target input. Suppose \(u(x)=mx+b\), \(v(x)=nx+d\), and \(u(a)=v(a)=c>0\). Their difference is \[ u(x)-v(x)=(m-n)x+(b-d). \] Because \(u(a)-v(a)=0\), this linear expression equals \((m-n)(x-a)\). After rationalization and cancellation, the quotient becomes \[ \frac{m-n}{\sqrt{u(x)}+\sqrt{v(x)}}. \] As \(x\) approaches \(a\), both square roots approach \(\sqrt c\), so the limit is \((m-n)/(2\sqrt c)\). This argument explains exactly why conjugate multiplication works in this setting: the difference of the radicands supplies the factor that cancels the original denominator.
Worked Examples
Worked Example: A Single Square Root and a Constant
Evaluate \[ \lim_{x\to 4}\frac{\sqrt{x+5}-3}{x-4}. \]
Step 1: Substitute directly. The numerator approaches \(\sqrt{4+5}-3=3-3=0\), and the denominator approaches \(4-4=0\). This is \(0/0\), so direct substitution does not evaluate the limit.
Step 2: Multiply by the conjugate. The conjugate of \(\sqrt{x+5}-3\) is \(\sqrt{x+5}+3\): \[ \frac{\sqrt{x+5}-3}{x-4} \cdot \frac{\sqrt{x+5}+3}{\sqrt{x+5}+3} = \frac{(x+5)-9}{(x-4)(\sqrt{x+5}+3)}. \]
Step 3: Factor and cancel. The numerator is \(x+5-9=x-4\). Thus, for \(x\ne4\), \[ \frac{(x+5)-9}{(x-4)(\sqrt{x+5}+3)} = \frac{x-4}{(x-4)(\sqrt{x+5}+3)} = \frac{1}{\sqrt{x+5}+3}. \]
Step 4: Evaluate. The simplified expression is defined and continuous at \(x=4\), so \[ \lim_{x\to4}\frac{1}{\sqrt{x+5}+3} = \frac{1}{\sqrt{4+5}+3} = \frac{1}{3+3} = \frac16. \]
Check: Multiplying the simplified expression by the canceled factor gives \((x-4)/((x-4)(\sqrt{x+5}+3))\), the rationalized original expression whenever \(x\ne4\). The limit is therefore \(1/6\).
Worked Example: Two Square Roots with Linear Radicands
Evaluate \[ \lim_{x\to2}\frac{\sqrt{x+3}-\sqrt{2x+1}}{x-2}. \]
Step 1: Check substitution. At \(x=2\), the radicands are \(2+3=5\) and \(2(2)+1=5\). The numerator approaches \(\sqrt5-\sqrt5=0\), and the denominator approaches \(0\), giving \(0/0\).
Step 2: Multiply by the conjugate. The conjugate of \(\sqrt{x+3}-\sqrt{2x+1}\) is \(\sqrt{x+3}+\sqrt{2x+1}\): \[ \frac{\sqrt{x+3}-\sqrt{2x+1}}{x-2} \cdot \frac{\sqrt{x+3}+\sqrt{2x+1}}{\sqrt{x+3}+\sqrt{2x+1}} = \frac{(x+3)-(2x+1)} {(x-2)(\sqrt{x+3}+\sqrt{2x+1})}. \]
Step 3: Simplify the radicand difference. \[ (x+3)-(2x+1)=x+3-2x-1=2-x=-(x-2). \] For \(x\ne2\), cancel \(x-2\): \[ \frac{-(x-2)}{(x-2)(\sqrt{x+3}+\sqrt{2x+1})} = -\frac{1}{\sqrt{x+3}+\sqrt{2x+1}}. \]
Step 4: Evaluate. \[ \lim_{x\to2}-\frac{1}{\sqrt{x+3}+\sqrt{2x+1}} = -\frac{1}{\sqrt5+\sqrt5} = -\frac{1}{2\sqrt5}. \]
Check: The radicand difference is exactly \(-(x-2)\), so the negative sign remains after cancellation. Substitution into the simplified expression gives the stated finite limit, with a positive denominator \(2\sqrt5\).
Worked Example: A Constant Square Root and a Variable Square Root
Evaluate \[ \lim_{x\to1}\frac{\sqrt{3x+1}-2}{x-1}. \]
Step 1: Substitute directly. The numerator approaches \(\sqrt{3(1)+1}-2=\sqrt4-2=0\), while \(x-1\) approaches \(0\). Direct substitution gives \(0/0\).
Step 2: Multiply by the conjugate. The conjugate is \(\sqrt{3x+1}+2\): \[ \frac{\sqrt{3x+1}-2}{x-1} \cdot \frac{\sqrt{3x+1}+2}{\sqrt{3x+1}+2} = \frac{(3x+1)-4}{(x-1)(\sqrt{3x+1}+2)}. \]
Step 3: Factor and cancel. Since \((3x+1)-4=3x-3=3(x-1)\), for \(x\ne1\), \[ \frac{3(x-1)}{(x-1)(\sqrt{3x+1}+2)} = \frac{3}{\sqrt{3x+1}+2}. \]
Step 4: Evaluate. \[ \lim_{x\to1}\frac{3}{\sqrt{3x+1}+2} = \frac{3}{\sqrt{3(1)+1}+2} = \frac{3}{2+2} = \frac34. \]
Check: The difference of squares in the rationalized numerator is \(3x+1-4=3(x-1)\), confirming both the factor and the coefficient \(3\). Substitution into the simplified expression gives \(3/4\).
A Common Cancellation Mistake
Cancellation applies to factors, not to individual terms that happen to appear in a sum or difference. Consider the incorrect line \[ \frac{\sqrt{x+5}-3}{x-4} \quad\longrightarrow\quad \frac{\sqrt{x+5}}{1}. \] This cancels \(x-4\) against pieces of the numerator even though \(x-4\) is not a factor of \(\sqrt{x+5}-3\). The conclusion is wrong: the proposed right-hand expression approaches \(3\) at \(x=4\), while the correct limit is \(1/6\).
The correction is to multiply by the conjugate first. Only after using the difference-of-squares identity do we find an actual factor \(x-4\): \[ (\sqrt{x+5}-3)(\sqrt{x+5}+3)=x+5-9=x-4. \] That factor can then be canceled with the denominator for \(x\ne4\).
Why This Technique Matters
Rationalizing is more than a way to remove a radical from a calculation. It changes the expression into an equivalent one whose structure is easier to inspect. The difference \(u(x)-v(x)\) often makes the shared factor visible, and canceling that factor turns an indeterminate substitution into an ordinary evaluation.
For a limit, the original and simplified expressions need to agree for inputs close to the target, not necessarily at the target itself. In these examples the original denominator is zero at the target, so the original quotient is not defined there. The cancellation is still legitimate because it gives an equal expression at nearby inputs where the factor is nonzero. After cancellation, direct substitution evaluates the limit of that equivalent expression.
Before finishing a problem, check three things: the conjugate has the opposite sign between the same terms; the difference of squares is calculated with the correct order; and the factor is canceled only after it has been factored and only for inputs where it is nonzero.
Check Your Understanding
For each limit, identify the conjugate and show the factorization that permits cancellation before evaluating.
- Evaluate \(\displaystyle\lim_{x\to3}\frac{\sqrt{x+6}-3}{x-3}\).
- Evaluate \(\displaystyle\lim_{x\to1}\frac{\sqrt{2x+7}-3}{x-1}\).
- In a limit with numerator \(\sqrt{u(x)}-\sqrt{v(x)}\), what expression is its conjugate, and what does their product equal?
- Why can a shared factor \(x-a\) be canceled when evaluating a limit as \(x\to a\), even though the factor is zero at \(a\)?
- Explain why canceling a term from a sum without first factoring is not valid.
