Why the Absolute Value Matters
A polynomial inside a square root often grows like a power of its input. Factoring out a suitable power can reveal the behavior of the radical and make a limit straightforward to evaluate. The crucial detail is that taking the square root of an even power does not always give the input power itself: it gives an absolute value.
That distinction matters especially when the input approaches negative infinity. For example, \(\sqrt{x^2}\) is not \(x\) for every real \(x\); it is \(|x|\). If a limit contains a denominator involving \(x\), losing this absolute value can reverse the answer’s sign.
Factoring an Even Power
Suppose \(P\) is a polynomial of even degree \(2m\), where \(m\) is a positive integer, and its leading coefficient is \(a>0\). For nonzero \(x\), factor \(x^{2m}\) from the polynomial inside the square root. The basic identity is:
This factorization applies wherever \(P(x)\ge 0\), so that the real square root is defined. Since \(P(x)/x^{2m}\to a>0\) at either end of the real line, the radicand is positive for all sufficiently large positive or negative \(x\). Thus, for a polynomial of even degree with positive leading coefficient, its square root is defined sufficiently far out in both directions.
The factor \(P(x)/x^{2m}\) tends to the leading coefficient \(a\): dividing each term of \(P\) by \(x^{2m}\) makes every lower-degree term tend to zero. When taking the square root of this remaining factor, the usual continuity of the square-root function at a positive number gives a limit of \(\sqrt a\).
A General Limit Rule
The factoring identity gives a useful rule for quotients. Let \(P\) have degree \(2m\) and leading coefficient \(a>0\), and let \(Q\) have degree \(m\) and leading coefficient \(c\ne 0\). Then the limits are:
These conclusions require the degrees and leading coefficients just stated. The radicand must be nonnegative for inputs sufficiently far in the direction of the limit; the assumptions \(a>0\) and even degree ensure this here. Also, \(Q(x)\) is nonzero for all sufficiently large \(|x|\), since its leading term is \(cx^m\).
Why the Rule Works
To see the sign and magnitude separately, divide both the numerator and denominator by \(x^m\), taking care not to replace the square-root factor \(|x|^m\) by \(x^m\). For sufficiently large \(|x|\),
As \(x\to+\infty\), the first factor tends to \(1\). As \(x\to-\infty\), it is \((-1)^m\): when \(m\) is even, \(x^m\) is positive; when \(m\) is odd, \(x^m\) is negative. Meanwhile, \(P(x)/x^{2m}\to a\) and \(Q(x)/x^m\to c\). The denominator’s normalized limit is nonzero, so the quotient limit law applies. Combining these limits proves the rule in both directions.
Worked Examples
Worked Example: A Quotient at Positive Infinity
Evaluate \(\displaystyle\lim_{x\to+\infty}\frac{\sqrt{9x^2+6x+5}}{3x+1}\).
The polynomial under the square root has degree \(2\), so factor out \(x^2\). Since \(x\to+\infty\), \(|x|=x\):
As \(x\to+\infty\), \(6/x\to0\), \(5/x^2\to0\), and \(1/x\to0\). Therefore, the numerator tends to \(\sqrt9=3\) and the denominator tends to \(3\). The limit is \(3/3=1\).
Check: The factored expression tends to \(3/3=1\), and the degree-\(2\) numerator and degree-\(1\) denominator have matching square-root growth, consistent with a finite, nonzero limit.
Worked Example: A Quotient at Negative Infinity
Evaluate \(\displaystyle\lim_{x\to-\infty}\frac{\sqrt{9x^2+6x+5}}{3x+1}\).
The factorization inside the radical is the same, but now \(|x|=-x\). Factor \(x^2\) from the radicand and \(x\) from the denominator:
As \(x\to-\infty\), \(|x|/x=-1\). The square-root factor tends to \(3\), and the denominator factor tends to \(3\). Hence the limit is \((-1)(3/3)=-1\).
Check: In the factored expression, the first factor is exactly \(-1\) for negative \(x\), while the remaining quotient tends to \(1\). Their product tends to \(-1\).
Worked Example: An Even Power at Negative Infinity
Evaluate \(\displaystyle\lim_{x\to-\infty}\frac{\sqrt{25x^4-10x^3+7x^2+1}}{5x^2-2x}\).
The polynomial under the radical has degree \(4\), so factor out \(x^4\). Because \(\sqrt{x^4}=|x^2|=x^2\), and \(x^2\) is nonnegative, this factor does not introduce a negative sign:
As \(x\to-\infty\), the terms \(1/x\), \(1/x^2\), and \(1/x^4\) all tend to zero. Thus the numerator tends to \(\sqrt{25}=5\), and the denominator tends to \(5\). The limit is \(5/5=1\).
Check: The final factored quotient tends directly to \(5/5=1\). The fourth power produced the factor \(x^2\), which is positive on both sides of zero, so no sign change is expected at negative infinity.
A Common Mistake
A frequent error is to use \(\sqrt{x^2}=x\) even when \(x\) is negative. For the second example, that mistake gives:
The first equality is incorrect for negative \(x\): \(\sqrt{x^2}=|x|=-x\), not \(x\). Keeping the absolute value gives:
The correction is simple but essential: factor out an even power inside the radical, then write its square root as an absolute value before simplifying.
When the Radicand Has Odd Degree
The same factoring identity can still help with an odd-degree polynomial, but the remaining factor is different. For example, a cubic can be written by factoring out \(x^2\), leaving a factor of degree one inside the square root. Before evaluating a real limit, check whether the radicand is nonnegative for inputs sufficiently far in the direction under consideration. A polynomial of odd degree with positive leading coefficient is eventually positive at positive infinity but eventually negative at negative infinity; in the latter direction, its real square root is not defined for sufficiently large negative inputs.
This domain check is part of the method, not an optional final detail. Factoring identifies the behavior, while the requirement \(P(x)\ge0\) determines whether the real-valued expression exists near the limit at all.
How to Use the Method
Confirm that the polynomial under the square root is nonnegative for inputs sufficiently far toward the limit.
For a degree-\(2m\) polynomial, factor out \(x^{2m}\) and divide each term by that power.
Replace \(\sqrt{x^{2m}}\) with \(|x|^m\); at negative infinity, determine its ratio to any remaining power of \(x\).
Terms with negative powers of \(x\) tend to zero, and the square-root factor tends to the square root of its positive limiting value.
Why This Method Matters
Factoring makes the scale and sign of a radical visible without needing to manipulate the expression by another method. In particular, it explains why two limits can have equal magnitudes but opposite signs: the absolute value from the square root is positive, while a denominator involving an odd power of \(x\) changes sign at negative infinity.
The key habit is to write \(\sqrt{x^{2m}}=|x|^m\) before canceling powers. Then check the direction of the limit, the sign of the remaining factors, and whether the radical is defined sufficiently far along that direction.
Check Your Understanding
Use factoring and retain the absolute value until the direction of the limit is clear.
- Evaluate \(\displaystyle\lim_{x\to+\infty}\frac{\sqrt{16x^2+4x+9}}{4x+2}\).
- Evaluate \(\displaystyle\lim_{x\to-\infty}\frac{\sqrt{16x^2+4x+9}}{4x+2}\).
- Why does \(\sqrt{x^4}=x^2\) for both positive and negative \(x\), while \(\sqrt{x^2}\) need not equal \(x\)?
- State the degree and leading-coefficient conditions for the quotient rule in this tutorial.
- Is the real-valued function \(\sqrt{x^3+2x}\) defined for all sufficiently large negative \(x\)? Explain using the sign of its radicand.