Why Conjugates Help with Radical Limits
A quotient involving square roots may approach zero over zero when its numerator is a difference of radicals that agree at the endpoint. Direct substitution then gives no value for the limit. In many such cases, multiplying by a conjugate changes the radical difference into an ordinary algebraic difference that has a factor in common with the denominator.
L'Hôpital's Rule can also evaluate these limits, provided its hypotheses are satisfied. Comparing the methods is useful: conjugate multiplication often exposes the factor causing the indeterminate form, while L'Hôpital's Rule replaces the quotient with one involving derivatives. Both methods must preserve the original limit and be applied under valid conditions.
The Basic Conjugate Calculation
Suppose a numerator has the form \(\sqrt{A(x)}-\sqrt{B(x)}\), where both radicands are nonnegative near the endpoint. Multiplication by the conjugate gives
The radicals disappear from the product. If \(A(x)-B(x)\) has a factor that also appears in the denominator, cancellation may reduce the quotient to an expression with a nonzero denominator at the endpoint. Cancellation is valid at nearby points where the factor is nonzero; it does not claim that the original quotient is defined at the endpoint.
For comparison, L'Hôpital's Rule applies to a quotient \(f(x)/g(x)\) near a finite endpoint \(a\) when \(f(a)\) and \(g(a)\) tend to zero, the functions are differentiable on a punctured neighborhood of \(a\), and \(g'(x)\ne 0\) there. If the limit of \(f'(x)/g'(x)\) exists (finite or infinite), then the limit of \(f(x)/g(x)\) equals it. For a one-sided limit, the conditions and the derivative quotient are considered on the corresponding side. The radical expressions must be defined on the relevant neighborhood as well.
Substitution producing \(0/0\) is a signal that a method may be useful, not a conclusion about the limit. For these examples, the resulting algebra or derivative quotient will establish the limit.
Worked Example: One Radical Difference over a Linear Factor
Worked Example: One Radical Difference over a Linear Factor
Evaluate \(\displaystyle\lim_{x\to 4}\frac{\sqrt{x+5}-3}{x-4}\).
At \(x=4\), the numerator is \(\sqrt{9}-3=0\), and the denominator is \(4-4=0\). Near \(4\), \(x+5>0\), so the radical is defined. Multiply by the conjugate of the numerator:
The cancellation is valid for \(x\ne4\) near the endpoint. The reduced expression is continuous at \(4\), so
Check with L'Hôpital's Rule. The numerator and denominator tend to zero, and both are differentiable near \(4\). The denominator derivative is \(1\), which is nonzero. Thus
Both methods give the same answer. The conjugate calculation displays the factor \(x-4\) that caused the indeterminate form; L'Hôpital's Rule reaches the answer directly by differentiating.
Why the Conjugate Method Works
The method is not an approximation. It follows from an exact algebraic identity at every nearby point where the expressions are defined. For two square roots, that identity converts a difference of roots into a difference of radicands. If this new difference contains the denominator's vanishing factor, cancellation leaves an equivalent quotient away from the endpoint.
Taking \(u=\sqrt{A(x)}\) and \(v=\sqrt{B(x)}\) gives \(u^2=A(x)\) and \(v^2=B(x)\), so the product is \(A(x)-B(x)\). The method is especially effective when \(A(x)-B(x)\) is simpler than the original radical difference and shares a factor with the denominator. If those features are absent, conjugation may not simplify the limit; L'Hôpital's Rule may be more convenient if its conditions hold.
Worked Example: Two Different Radicals in the Numerator
Worked Example: Two Different Radicals in the Numerator
Evaluate \(\displaystyle\lim_{x\to0}\frac{\sqrt{1+3x}-\sqrt{1-x}}{x}\).
At \(x=0\), the numerator is \(1-1=0\), and the denominator is zero. Both radicands are positive for \(x\) in a sufficiently small neighborhood of zero. Multiply by the conjugate of the entire numerator:
Taking the limit of the reduced expression gives
For a check, differentiate the numerator and denominator. The numerator derivative is \(\frac{3}{2\sqrt{1+3x}}+\frac{1}{2\sqrt{1-x}}\): the second term is positive because differentiating \(-\sqrt{1-x}\) gives \(+\frac{1}{2\sqrt{1-x}}\). The denominator derivative is \(1\). Hence
The two approaches agree. Here the conjugate step makes the subtraction of the radicands explicit, while L'Hôpital's Rule requires careful attention to the derivative sign.
Worked Example: A Quotient of Two Radical Differences
Worked Example: A Quotient of Two Radical Differences
Evaluate \(\displaystyle\lim_{x\to0}\frac{\sqrt{4+x}-2}{\sqrt{9+x}-3}\).
The numerator and denominator both tend to zero. Rationalize each difference separately:
For \(x\ne0\) near zero, substitute these equal expressions into the quotient and cancel \(x\):
The reduced denominator tends to \(2+2=4\), so the limit is
Check using L'Hôpital's Rule. The derivative of the numerator is \(1/(2\sqrt{4+x})\); the derivative of the denominator is \(1/(2\sqrt{9+x})\). The latter is nonzero near zero. Therefore
This example illustrates a useful choice: rationalizing the two differences individually reveals a common factor \(x\), even though there is no single radical difference in the original numerator to rationalize.
Choosing a Method and Avoiding Errors
For a finite-endpoint radical quotient of type \(0/0\), first check the radicands and the exact form of the numerator and denominator. Then look for a conjugate that turns a radical difference into a factorable difference of radicands. If this exposes a common factor, conjugation is often the clearest route. If the derivatives are simpler and L'Hôpital's hypotheses hold, applying the rule may be shorter. You can use the other method as a check when practical.
A common error is to subtract the radicands incorrectly when multiplying by a conjugate. In the second worked example, the relevant difference is
not \(2x\). The minus sign before the second parentheses changes the sign of both terms inside them. As a result, an incorrect subtraction would change the final limit.
Another pitfall is treating cancellation at the endpoint as though the original quotient were defined there. In these examples, the factor is cancelled only for \(x\ne a\), where \(a\) is the endpoint. Limits depend on values arbitrarily near \(a\), not on assigning a value to the original quotient at \(a\). A third pitfall is applying L'Hôpital's Rule merely because radicals appear: the original quotient must have the required indeterminate form, and the differentiability and denominator-derivative conditions must hold.
Check Your Understanding
For each question, decide which method is most useful and justify the steps that determine the limit.
- Evaluate \(\displaystyle\lim_{x\to3}\frac{\sqrt{x+6}-3}{x-3}\) by conjugate multiplication, then check it with L'Hôpital's Rule.
- In rationalizing \(\sqrt{2+5x}-\sqrt{2-x}\), what difference of radicands results after multiplying by the conjugate?
- Evaluate \(\displaystyle\lim_{x\to0}\frac{\sqrt{1+4x}-1}{x}\) using either method, and state why the method applies.
- When cancelling a factor that vanishes at the endpoint, on which nearby values is that cancellation valid?
- What must be checked before applying L'Hôpital's Rule to a radical quotient?
