Tutorials › Calculus II › Radical Sequence Limits by Conjugation

Home  /  Tutorials  /  Sequence Limits

Sequence Limits · Tutorial 507 of 998

Radical Sequence Limits by Conjugation

Use conjugates to simplify radical differences, compare powers of the index, and determine whether the sequence has a finite limit.

Beginner 10 min read

What You'll Learn

  • Apply the difference-of-squares identity to radical sequence terms
  • Identify when conjugation is valid and when a finite limit is plausible
  • Compare powers of the index after simplifying a radical difference
  • Prove a general limit formula for square roots with matching leading terms
  • Check answers and avoid common errors with signs and denominators

Why Conjugation Helps

A difference of square roots can be difficult to evaluate by looking at its two terms separately. For example, both \(\sqrt{n^2+5n+6}\) and \(n\) grow without bound, yet their difference approaches a finite value. The useful information is hidden in the difference.

Multiplying by the conjugate exposes that information. The resulting numerator is a difference of squares, so the radical terms can be replaced by a simpler polynomial expression. The remaining limit can then be analyzed by comparing powers of \(n\), as in the earlier material on rational sequences and dominant powers.

Key idea: To simplify a difference of square roots, multiply and divide by the corresponding sum of square roots. This is useful when the original terms have similar leading growth.

The Conjugate Identity

For expressions \(A\) and \(B\), the conjugate of \(A-B\) is \(A+B\). The difference-of-squares identity gives

$$ (A-B)(A+B)=A^2-B^2. $$

Consequently, whenever \(A+B\ne 0\),

$$ A-B=\frac{A^2-B^2}{A+B}. $$

For square roots of nonnegative quantities, this gives

$$ \sqrt{X_n}-\sqrt{Y_n} = \frac{X_n-Y_n}{\sqrt{X_n}+\sqrt{Y_n}}, $$

provided the denominator is nonzero. In sequence problems, it is enough for the radicands to be nonnegative and the denominator positive for all sufficiently large \(n\). A finite number of initial terms does not affect whether a sequence converges or what its limit is.

The identity is exact, not an approximation. Its advantage is that \(X_n-Y_n\) may be much simpler than the original difference of roots.

A Useful Fact About Square-Root Limits

After conjugating, we often divide a radicand by a power of \(n\). To evaluate the resulting square-root terms, we need a basic fact. Suppose \(u_n\ge 0\) and \(u_n\to a\), where \(a>0\). Then

$$ \left|\sqrt{u_n}-\sqrt a\right| = \frac{|u_n-a|}{\sqrt{u_n}+\sqrt a} \le \frac{|u_n-a|}{\sqrt a}. $$

The inequality holds because \(\sqrt{u_n}\ge 0\), so the denominator is at least \(\sqrt a\). Since \(u_n\to a\), the right-hand side tends to zero. Therefore \(\sqrt{u_n}\to\sqrt a\). This argument justifies the square-root limits used below without relying on a limit rule for continuous functions.

Worked Example: A Difference with a Finite Nonzero Limit

Worked Example: A Difference with a Finite Nonzero Limit

Find the limit of \(a_n=\sqrt{n^2+5n+6}-n\), where \(n\in\mathbb N\).

The radicand is positive for every positive integer \(n\), and the conjugate denominator below is positive. Multiply by the conjugate and simplify:

$$ \begin{aligned} a_n &=\frac{\left(\sqrt{n^2+5n+6}-n\right) \left(\sqrt{n^2+5n+6}+n\right)} {\sqrt{n^2+5n+6}+n}\\ &=\frac{n^2+5n+6-n^2}{\sqrt{n^2+5n+6}+n}\\ &=\frac{5n+6}{\sqrt{n^2+5n+6}+n}. \end{aligned} $$

Divide the numerator and denominator by \(n\), which is positive:

$$ a_n = \frac{5+6/n}{\sqrt{1+5/n+6/n^2}+1}. $$

Here \(1+5/n+6/n^2\to 1\), so the square-root fact above gives \(\sqrt{1+5/n+6/n^2}\to1\). The numerator tends to \(5\), and the denominator tends to \(2\). The quotient limit law therefore gives

$$ \lim_{n\to\infty}a_n=\frac{5}{2}. $$

Check: The conjugation identity is exact, and the simplified quotient has limiting numerator \(5\) and limiting denominator \(2\). Thus the original difference tends to \(5/2\), not to zero.

Worked Example: A Difference That Tends to Zero

Worked Example: A Difference That Tends to Zero

Find the limit of \(b_n=\sqrt{n^2+1}-n\).

Conjugating gives an exact expression:

$$ \begin{aligned} b_n &=\frac{(\sqrt{n^2+1}-n)(\sqrt{n^2+1}+n)} {\sqrt{n^2+1}+n}\\ &=\frac{n^2+1-n^2}{\sqrt{n^2+1}+n}\\ &=\frac{1}{\sqrt{n^2+1}+n}. \end{aligned} $$

For \(n\ge1\), \(\sqrt{n^2+1}\ge n\), so the denominator is at least \(2n\). Hence

$$ 0<b_n\le\frac{1}{2n}. $$

Because \(1/(2n)\to0\), these inequalities show that \(b_n\to0\). Equivalently, dividing the exact expression by \(n\) gives \(b_n=1/\bigl(n(\sqrt{1+1/n^2}+1)\bigr)\), whose denominator grows without bound.

Check: The conjugated expression is positive and no larger than \(1/(2n)\), so it approaches zero from above. This agrees with the original expression: \(\sqrt{n^2+1}\) is greater than \(n\), but their difference becomes small.

Worked Example: Two Roots with the Same Leading Growth

Worked Example: Two Roots with the Same Leading Growth

Find the limit of \(c_n=\sqrt{4n^2+3n+1}-\sqrt{4n^2-n+2}\).

For positive integers \(n\), both radicands are positive. Multiply by the sum of the roots:

$$ \begin{aligned} c_n &=\frac{(4n^2+3n+1)-(4n^2-n+2)} {\sqrt{4n^2+3n+1}+\sqrt{4n^2-n+2}}\\ &=\frac{4n-1} {\sqrt{4n^2+3n+1}+\sqrt{4n^2-n+2}}. \end{aligned} $$

Divide numerator and denominator by \(n\):

$$ c_n = \frac{4-1/n} {\sqrt{4+3/n+1/n^2}+\sqrt{4-1/n+2/n^2}}. $$

Each radicand in the denominator tends to \(4\), so each square root tends to \(2\). The numerator tends to \(4\), and the denominator tends to \(2+2=4\). Thus

$$ \lim_{n\to\infty}c_n=\frac{4}{4}=1. $$

Check: The exact numerator after conjugation is \(4n-1\), while the denominator divided by \(n\) tends to \(4\). The transformed quotient therefore has the same limit \(1\) as the original difference.

A General Formula and Its Derivation

The preceding example follows a general pattern. Suppose \(\alpha>0\), and \(\beta,\gamma,\delta,\eta\) are fixed real numbers. Assume the radicands are nonnegative for all sufficiently large \(n\). Then

Theorem: If the two square roots are defined for all sufficiently large \(n\), then \(\sqrt{\alpha n^2+\beta n+\gamma}-\sqrt{\alpha n^2+\delta n+\eta}\) tends to \((\beta-\delta)/(2\sqrt{\alpha})\).

To prove the formula, conjugate the difference. The difference of the radicands is \((\beta-\delta)n+(\gamma-\eta)\), so

$$ \begin{aligned} &\sqrt{\alpha n^2+\beta n+\gamma} -\sqrt{\alpha n^2+\delta n+\eta}\\ &\quad= \frac{(\beta-\delta)n+(\gamma-\eta)} {\sqrt{\alpha n^2+\beta n+\gamma} +\sqrt{\alpha n^2+\delta n+\eta}}. \end{aligned} $$

Since \(n>0\), divide the numerator and denominator by \(n\). The expression becomes

$$ \frac{(\beta-\delta)+(\gamma-\eta)/n} {\sqrt{\alpha+\beta/n+\gamma/n^2} +\sqrt{\alpha+\delta/n+\eta/n^2}}. $$

The radicands in the denominator each tend to \(\alpha>0\). The square-root fact proved earlier shows that each root tends to \(\sqrt{\alpha}\). The numerator tends to \(\beta-\delta\), while the denominator tends to \(2\sqrt{\alpha}\), which is nonzero. Applying the quotient limit law proves the stated limit.

This also explains why matching leading terms matter. The two roots each have leading growth \(\sqrt{\alpha}\,n\). Conjugation cancels the matching \(\alpha n^2\) terms in the numerator, leaving terms of order at most \(n\); after dividing numerator and denominator by \(n\), a finite limit remains.

Worked Example: Comparing Higher Powers

Worked Example: Comparing Higher Powers

Find the limit of \(d_n=\sqrt{9n^4+2n^2}-3n^2\).

The radicand is positive. Conjugating cancels the \(9n^4\) terms:

$$ \begin{aligned} d_n &=\frac{(\sqrt{9n^4+2n^2}-3n^2)(\sqrt{9n^4+2n^2}+3n^2)} {\sqrt{9n^4+2n^2}+3n^2}\\ &=\frac{9n^4+2n^2-9n^4} {\sqrt{9n^4+2n^2}+3n^2}\\ &=\frac{2n^2}{\sqrt{9n^4+2n^2}+3n^2}. \end{aligned} $$

Divide the numerator and denominator by \(n^2\), which is positive:

$$ d_n=\frac{2}{\sqrt{9+2/n^2}+3}. $$

The radicand tends to \(9\), so the denominator tends to \(3+3=6\). Therefore

$$ \lim_{n\to\infty}d_n=\frac{2}{6}=\frac{1}{3}. $$

Check: The exact numerator after conjugation is \(2n^2\). Dividing by the same power \(n^2\) in the denominator leaves the quotient \(2/(\sqrt{9+2/n^2}+3)\), which tends to \(1/3\).

Choosing the Right Power and Avoiding Errors

After conjugation, inspect the largest powers remaining in the numerator and denominator. Divide both by the same power so that each part has a finite limit, or so that the growth of the denominator is clear. For instance, in the last example both remaining terms have order \(n^2\), so dividing by \(n^2\) is natural.

A frequent mistake is to treat \(\sqrt{n^2+5n+6}-n\) as if it were approximately zero simply because the leading terms under the root and outside it look alike. That misses the contribution from the \(5n\) term. The exact calculation is

$$ \sqrt{n^2+5n+6}-n = \frac{5n+6}{\sqrt{n^2+5n+6}+n} \longrightarrow \frac{5}{2}, $$

not zero. Another mistake is to forget that the conjugate denominator is a sum. Replacing it with a difference does not produce the difference-of-squares identity and changes the expression. Keep the identity visible: multiply by the sum, and simplify the product as \(A^2-B^2\).

Conjugation is especially effective when the leading terms of the roots match. If their leading coefficients differ, the difference may grow rather than approach a finite number. For example, \(\sqrt{4n^2}-\sqrt{n^2}=2n-n=n\), which does not have a finite limit. Simplification is still useful, but it does not guarantee convergence.

Check Your Understanding

For each problem, identify the conjugate and the power of \(n\) that is useful for simplifying the limit.

  1. Find \(\lim_{n\to\infty}\bigl(\sqrt{n^2+4n+3}-n\bigr)\).
  2. Find \(\lim_{n\to\infty}\bigl(\sqrt{n^2+4}-n\bigr)\).
  3. Find \(\lim_{n\to\infty}\bigl(\sqrt{4n^2+2n}-\sqrt{4n^2-6n}\bigr)\).
  4. In the general formula, why must \(\alpha>0\) for the stated finite limit argument?
  5. Explain why observing that two terms both grow without bound does not determine the limit of their difference.

Previous

Next

Back to Tutorials