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Algebraic Limit Techniques · Tutorial 37 of 999

Nested Radicals and Conjugate Multiplication

Use the conjugate of the outer radical to expose a factor that matches the denominator, then evaluate the simplified limit.

Beginner 9 min read

What You'll Learn

  • Identify the correct conjugate when a square root contains another radical.
  • Use a difference-of-squares identity to expose an inner-radical difference.
  • State when the resulting quotient has a finite limit.
  • Work through nested-radical limits with exact values.
  • Recognize why cancelling terms inside a sum or difference is invalid.
  • Check the simplified expression before evaluating the limit.

Why Nested Radicals Need a Careful Conjugate

A limit involving a nested square root can produce a zero-over-zero form even when its value is finite. The useful algebraic move is to focus on the outer radical: multiply by the conjugate of the entire numerator, not by a conjugate chosen from only part of the expression. The resulting difference of squares may reveal the inner-radical difference that matches the denominator.

For example, in \(\sqrt{7+\sqrt{x}}-3\), the two terms being subtracted are \(\sqrt{7+\sqrt{x}}\) and \(3\). Their conjugate is \(\sqrt{7+\sqrt{x}}+3\). Multiplying by it removes the outer square root through a difference of squares:

$$ \left(\sqrt{7+\sqrt{x}}-3\right) \left(\sqrt{7+\sqrt{x}}+3\right) = 7+\sqrt{x}-9 = \sqrt{x}-2. $$

The expression on the right is an inner-radical difference. If the original denominator contains \(\sqrt{x}-2\), the multiplication has exposed a factor that can be cancelled.

The Algebra Behind the Method

Definition: The conjugate of a binomial \(A-B\) is \(A+B\); the conjugate of \(A+B\) is \(A-B\). Their product is the difference of squares, \((A-B)(A+B)=A^2-B^2\). For a nested radical, identify \(A\) and \(B\) as the two complete terms on either side of the subtraction.

Here is a general form of the method. Suppose \(g\) is continuous near \(x=a\), \(g(x)\geq 0\) near \(a\), and \(A+\sqrt{g(a)}>0\). Consider the quotient

$$ \frac{\sqrt{A+\sqrt{g(x)}}-\sqrt{A+\sqrt{g(a)}}} {\sqrt{g(x)}-\sqrt{g(a)}}. $$

At \(x=a\), both numerator and denominator are zero. For nearby \(x\) where the denominator is nonzero, multiply the numerator and denominator by the conjugate of the numerator. This gives

$$ \begin{aligned} &\frac{\sqrt{A+\sqrt{g(x)}}-\sqrt{A+\sqrt{g(a)}}} {\sqrt{g(x)}-\sqrt{g(a)}}\\[4pt] &\quad= \frac{A+\sqrt{g(x)}-\bigl(A+\sqrt{g(a)}\bigr)} {\bigl(\sqrt{g(x)}-\sqrt{g(a)}\bigr) \bigl(\sqrt{A+\sqrt{g(x)}}+\sqrt{A+\sqrt{g(a)}}\bigr)}\\[4pt] &\quad= \frac{\sqrt{g(x)}-\sqrt{g(a)}} {\bigl(\sqrt{g(x)}-\sqrt{g(a)}\bigr) \bigl(\sqrt{A+\sqrt{g(x)}}+\sqrt{A+\sqrt{g(a)}}\bigr)}\\[4pt] &\quad= \frac{1}{\sqrt{A+\sqrt{g(x)}}+\sqrt{A+\sqrt{g(a)}}}. \end{aligned} $$

The cancellation is valid for the nearby \(x\) under consideration because \(\sqrt{g(x)}-\sqrt{g(a)}\ne 0\) there. A limit concerns values near \(a\), not necessarily the expression’s value at \(a\). By continuity, as \(x\) approaches \(a\), the remaining denominator approaches \(2\sqrt{A+\sqrt{g(a)}}\), which is positive by assumption. Therefore:

Theorem: Under the conditions above, if the quotient is defined for \(x\) sufficiently near \(a\) with \(x\ne a\), then $$ \lim_{x\to a} \frac{\sqrt{A+\sqrt{g(x)}}-\sqrt{A+\sqrt{g(a)}}} {\sqrt{g(x)}-\sqrt{g(a)}} = \frac{1}{2\sqrt{A+\sqrt{g(a)}}}. $$ The key step is that conjugating the outer radical turns its numerator into the same inner-radical difference as the denominator.

Worked Examples

Worked Example: A Matching Inner-Radical Difference

Evaluate

$$ \lim_{x\to 4} \frac{\sqrt{7+\sqrt{x}}-3}{\sqrt{x}-2}. $$

Step 1: Check direct substitution. At \(x=4\), \(\sqrt{x}=2\), and \(\sqrt{7+2}=3\). Thus both numerator and denominator approach zero.

$$ \frac{\sqrt{7+\sqrt{4}}-3}{\sqrt{4}-2} = \frac{3-3}{2-2} = \frac{0}{0}. $$

Step 2: Multiply by the outer conjugate. The conjugate of \(\sqrt{7+\sqrt{x}}-3\) is \(\sqrt{7+\sqrt{x}}+3\). Multiply by this conjugate over itself:

$$ \begin{aligned} \frac{\sqrt{7+\sqrt{x}}-3}{\sqrt{x}-2} &= \frac{(\sqrt{7+\sqrt{x}}-3)(\sqrt{7+\sqrt{x}}+3)} {(\sqrt{x}-2)(\sqrt{7+\sqrt{x}}+3)}\\ &= \frac{7+\sqrt{x}-9} {(\sqrt{x}-2)(\sqrt{7+\sqrt{x}}+3)}\\ &= \frac{\sqrt{x}-2} {(\sqrt{x}-2)(\sqrt{7+\sqrt{x}}+3)}. \end{aligned} $$

Step 3: Cancel and evaluate. For \(x\ne4\) near \(4\), cancel the common factor \(\sqrt{x}-2\). The remaining expression is continuous at \(4\):

$$ \lim_{x\to4}\frac{1}{\sqrt{7+\sqrt{x}}+3} = \frac{1}{\sqrt{7+\sqrt{4}}+3} = \frac{1}{3+3} = \frac{1}{6}. $$

Check: The general result gives \(1/(2\sqrt{7+\sqrt{4}})=1/(2\cdot3)=1/6\), agreeing with the simplified limit.

Worked Example: An Inner Radical with a Shift

Evaluate

$$ \lim_{x\to1} \frac{\sqrt{2+\sqrt{x+3}}-2}{\sqrt{x+3}-2}. $$

Step 1: Confirm the zero-over-zero form. At \(x=1\), \(\sqrt{x+3}=\sqrt4=2\), so the outer radical is \(\sqrt{2+2}=2\). Both numerator and denominator are zero.

$$ \frac{\sqrt{2+\sqrt{1+3}}-2}{\sqrt{1+3}-2} = \frac{2-2}{2-2} = \frac{0}{0}. $$

Step 2: Conjugate the whole numerator. Its conjugate is \(\sqrt{2+\sqrt{x+3}}+2\). Multiplication gives

$$ \begin{aligned} \frac{\sqrt{2+\sqrt{x+3}}-2}{\sqrt{x+3}-2} &= \frac{2+\sqrt{x+3}-4} {(\sqrt{x+3}-2)(\sqrt{2+\sqrt{x+3}}+2)}\\ &= \frac{\sqrt{x+3}-2} {(\sqrt{x+3}-2)(\sqrt{2+\sqrt{x+3}}+2)}. \end{aligned} $$

Step 3: Cancel and substitute into the remaining expression. For \(x\ne1\) near \(1\), cancel the common factor:

$$ \lim_{x\to1}\frac{1}{\sqrt{2+\sqrt{x+3}}+2} = \frac{1}{\sqrt{2+\sqrt4}+2} = \frac{1}{2+2} = \frac{1}{4}. $$

Check: The outer radical tends to \(2\), so the conjugate denominator tends to \(2+2=4\). Its reciprocal is \(1/4\), as found.

Worked Example: A Nonlinear Expression Inside the Inner Radical

Evaluate

$$ \lim_{x\to2} \frac{\sqrt{13+\sqrt{x^2+5}}-4}{\sqrt{x^2+5}-3}. $$

Step 1: Substitute to see why simplification is needed. At \(x=2\), \(x^2+5=4+5=9\), so the inner square root is \(3\). The outer radicand is \(13+3=16\), and the outer square root is \(4\). Thus the quotient is initially \(0/0\).

$$ \frac{\sqrt{13+\sqrt{2^2+5}}-4}{\sqrt{2^2+5}-3} = \frac{\sqrt{13+3}-4}{3-3} = \frac{4-4}{0} = \frac{0}{0}. $$

Step 2: Use the outer conjugate. The conjugate of \(\sqrt{13+\sqrt{x^2+5}}-4\) is \(\sqrt{13+\sqrt{x^2+5}}+4\):

$$ \begin{aligned} \frac{\sqrt{13+\sqrt{x^2+5}}-4}{\sqrt{x^2+5}-3} &= \frac{13+\sqrt{x^2+5}-16} {(\sqrt{x^2+5}-3)(\sqrt{13+\sqrt{x^2+5}}+4)}\\ &= \frac{\sqrt{x^2+5}-3} {(\sqrt{x^2+5}-3)(\sqrt{13+\sqrt{x^2+5}}+4)}. \end{aligned} $$

Step 3: Cancel and find the limit. For \(x\ne2\) sufficiently near \(2\), the common factor can be cancelled. Then continuity allows substitution:

$$ \lim_{x\to2}\frac{1}{\sqrt{13+\sqrt{x^2+5}}+4} = \frac{1}{\sqrt{13+\sqrt{2^2+5}}+4} = \frac{1}{4+4} = \frac{1}{8}. $$

Check: The conjugate denominator tends to \(4+4=8\), so the simplified expression tends to \(1/8\), confirming the result.

Conditions and a Common Cancellation Error

The method depends on more than spotting two radicals. The expressions must be defined for \(x\) near the point of approach, and the conjugate must produce a factor that matches the original denominator. After cancellation, the remaining expression must have a limit that can be evaluated. The conjugate multiplication is an identity wherever the original expressions are defined; cancellation is performed only at nearby points where the cancelled factor is nonzero.

A common error is to cancel part of a sum or difference as though it were a factor. For instance, someone might look at

$$ \frac{\sqrt{7+\sqrt{x}}-3}{\sqrt{x}-2} $$

and claim that the numerator contains \(\sqrt{x}-2\), so the quotient is \(1\). It does not contain that factor in its displayed form: the numerator is a difference of an outer radical and \(3\), not a product with \(\sqrt{x}-2\). The claim also fails at \(x=9\), where the original quotient is

$$ \frac{\sqrt{7+3}-3}{3-2} = \frac{\sqrt{10}-3}{1} = \sqrt{10}-3, $$

which is not \(1\). The correction is to multiply by the outer conjugate first. That operation proves the identity with a factor \(\sqrt{x}-2\); only then is there a genuine common factor to cancel.

A Reliable Sequence of Steps

When a nested-radical limit has a zero numerator and denominator, use this sequence. It keeps the algebra tied to the structure of the expression.

1
Substitute first.
Evaluate the inner and outer radicals at the approach point. Confirm that direct substitution gives the form that needs simplification.
2
Choose the outer conjugate.
For a numerator of the form \(\sqrt{A+\sqrt{g(x)}}-c\), use \(\sqrt{A+\sqrt{g(x)}}+c\). Treat each side of the subtraction as one complete term.
3
Apply the difference of squares.
Expand as the square of the entire outer radical minus \(c^2\). Look for the resulting inner-radical difference to match a denominator factor.
4
Cancel only a common factor.
State that the factor is nonzero for the nearby points where cancellation is used. Do not cancel a term that appears only as part of a sum or difference.
5
Evaluate the remaining limit.
Once the simplified expression is continuous at the approach point, substitute and check the resulting value.

Why This Method Is Useful

Nested radicals can hide a simple difference inside a more complicated expression. Multiplying by the outer conjugate removes one layer of that structure and can expose exactly the difference needed for cancellation. This is why the choice of conjugate matters: it should correspond to the two terms in the outer subtraction.

The process also illustrates a general principle of limit simplification. An expression may be undefined at the point of approach, yet equal a simpler expression at nearby points. If the simpler expression has a limit, it determines the limit of the original expression. The original and simplified expressions need not have the same value at the approach point; cancellation is used on a punctured neighborhood, not at the point itself.

Always finish by checking the algebra and the conditions. Verify that the conjugate was applied to the complete numerator, that the difference of squares has the correct sign, that the factor cancelled really occurs in both numerator and denominator, and that the remaining denominator does not approach zero.

Check Your Understanding

For each limit, identify the outer conjugate and the inner-radical difference it should reveal before evaluating.

  1. Evaluate \(\displaystyle \lim_{x\to4}\frac{\sqrt{5+\sqrt{x}}-3}{\sqrt{x}-2}\).
  2. For \(\displaystyle \sqrt{6+\sqrt{x+3}}-\sqrt{8}\), what is the conjugate of the entire numerator?
  3. Evaluate \(\displaystyle \lim_{x\to1}\frac{\sqrt{3+\sqrt{x+3}}-\sqrt{5}}{\sqrt{x+3}-2}\).
  4. Explain why a term inside a sum or difference cannot be cancelled as though it were a common factor.
  5. State why cancellation in a limit simplification does not require the cancelled factor to be nonzero at the approach point itself.

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