Comparing Powers at Infinity
For polynomials, the degree of the leading term determines end behavior. The same basic comparison applies to ratios of powers with real exponents: the difference between the exponents determines how the magnitude changes. But at negative infinity, growth rate alone is not enough. A power may be undefined for negative inputs, and a defined power may be positive or negative.
This tutorial develops a precise method for ratios of the form \(c x^\alpha/(d x^\beta)\), where \(c\) and \(d\) are nonzero constants and \(\alpha,\beta\) are real exponents. The method has two parts: compare magnitudes, then determine the sign.
Real Powers on Positive Inputs
For \(x>0\), the real power \(x^\alpha\) is defined for every real number \(\alpha\). It is positive, and its magnitude increases without bound as \(x\to+\infty\) when \(\alpha>0\), equals \(1\) when \(\alpha=0\), and tends to \(0\) when \(\alpha<0\).
The exponent laws give \(x^\alpha/x^\beta=x^{\alpha-\beta}\) for \(x>0\). Thus, for nonzero \(c,d\), the ratio \(c x^\alpha/(d x^\beta)\) is \((c/d)x^{\alpha-\beta}\). Its limit at positive infinity is determined by whether \(\alpha-\beta\) is positive, zero, or negative.
Worked Example: A Positive-Infinity Limit With Fractional Exponents
Worked Example: A Positive-Infinity Limit With Fractional Exponents
Evaluate \(\displaystyle \lim_{x\to+\infty}\frac{6x^{7/3}}{9x^{11/6}}\).
For positive \(x\), all real powers are defined and positive. First simplify the coefficient: \(\frac{6}{9}=\frac{2}{3}\). Next subtract the denominator exponent from the numerator exponent: \(\frac{7}{3}-\frac{11}{6}=\frac{14}{6}-\frac{11}{6}=\frac{1}{2}\). Therefore \[ \frac{6x^{7/3}}{9x^{11/6}}=\frac{2}{3}x^{1/2}. \] Since \(x^{1/2}\to+\infty\) as \(x\to+\infty\), the positive coefficient preserves the sign and \[ \lim_{x\to+\infty}\frac{6x^{7/3}}{9x^{11/6}}=+\infty. \] To check, at \(x=4\) the original ratio is \(6\cdot 4^{7/3}/(9\cdot4^{11/6})=(2/3)4^{1/2}=4/3\), matching the simplified expression.
Why Subtracting Exponents Works
The comparison can be justified by factoring out the smaller power. For positive \(x\), the exponent law immediately gives the ratio as a single power. At negative inputs, the same magnitude comparison is available when the powers are real-valued, but signs must be handled separately.
This argument separates two questions that are easy to confuse: How large is the ratio? is answered by the exponent difference; Which direction does it go? is answered by its sign.
Real Powers at Negative Inputs
For \(x<0\), an arbitrary real exponent does not necessarily give a real number. Under the usual real-valued definition, \(x^{m/n}\) is real for negative \(x\) when \(m/n\) is in lowest terms and its denominator \(n\) is odd. Integer exponents are included, since their denominator is \(1\). For example, \(x^{1/3}\) and \(x^{2/3}\) are real for negative \(x\), but \(x^{1/2}\) is not.
When \(m/n\) is reduced and \(n\) is odd, the sign is determined by the numerator \(m\). If \(m\) is even, \(x^{m/n}\) is positive for negative \(x\); if \(m\) is odd, it is negative. The magnitude is \(|x|^{m/n}\). This is why the sign should be checked using the original powers rather than inferred from the exponent difference alone.
Worked Example: Integer Powers at Negative Infinity
Worked Example: Integer Powers at Negative Infinity
Evaluate \(\displaystyle \lim_{x\to-\infty}\frac{5x^4}{-2x^3}\).
Both powers are integer powers, so they are defined for every real \(x\ne 0\). Simplify the coefficient and subtract the exponents: \[ \frac{5x^4}{-2x^3}=-\frac{5}{2}x^{4-3}=-\frac{5}{2}x. \] As \(x\to-\infty\), \(x\) is negative with magnitude increasing without bound. Multiplication by \(-5/2\) makes the expression positive with magnitude increasing without bound. Hence \[ \lim_{x\to-\infty}\frac{5x^4}{-2x^3}=+\infty. \] For a sign check, if \(x=-10\), the original expression is \(5(10{,}000)/[-2(-1{,}000)]=50{,}000/2{,}000=25\), and the simplified expression is \(-\frac52(-10)=25\).
A Sign-and-Growth Rule at Negative Infinity
For integer powers, and more generally for rational powers that are real-valued on negative inputs, the magnitude of a ratio is governed by \(|x|^{\alpha-\beta}\). Its sign is the sign of the full ratio, including the coefficients and the signs of each power. If the exponent difference is positive, the magnitude grows without bound; the sign then determines whether the limit is \(+\infty\) or \(-\infty\). If the exponent difference is negative, the magnitude tends to zero from the side indicated by the sign, so the two-sided limit at negative infinity is \(0\).
When the exponents are rational with odd denominators in reduced form, this sign can be found by checking whether each numerator is even or odd. The next example shows why keeping that check separate is useful.
Worked Example: Fractional Powers at Negative Infinity
Worked Example: Fractional Powers at Negative Infinity
Evaluate \(\displaystyle \lim_{x\to-\infty}\frac{-3x^{5/3}}{2x^{2/3}}\).
The exponents \(5/3\) and \(2/3\) have odd denominators, so both powers are real for negative \(x\). Their magnitudes are \(|x|^{5/3}\) and \(|x|^{2/3}\). The sign of \(x^{5/3}\) is negative because its numerator \(5\) is odd; the sign of \(x^{2/3}\) is positive because its numerator \(2\) is even. Thus the fraction of powers is negative, and the coefficient \(-3/2\) makes the entire expression positive.
For its magnitude, subtract exponents: \[ \left|\frac{-3x^{5/3}}{2x^{2/3}}\right| =\frac{3}{2}|x|^{5/3-2/3} =\frac{3}{2}|x|. \] This tends to infinity. Therefore the signed limit is \(+\infty\). Equivalently, the quotient simplifies to \(-\frac32 x\), since \(x^{5/3}/x^{2/3}=x\) for negative \(x\) under these real-power definitions. As \(x\to-\infty\), \(-\frac32x\to+\infty\), confirming the result.
Worked Example: A Ratio That Tends to Zero
Worked Example: A Ratio That Tends to Zero
Evaluate \(\displaystyle \lim_{x\to-\infty}\frac{4x^2}{-8x^5}\).
Both exponents are integers, so the expression is real-valued for negative \(x\). Simplifying gives \[ \frac{4x^2}{-8x^5}=-\frac12 x^{2-5}=-\frac{1}{2x^3}. \] The exponent difference is \(-3\), so the magnitude is \(\frac12|x|^{-3}=\frac{1}{2|x|^3}\), which tends to zero. For negative \(x\), \(x^3<0\), so \(-1/(2x^3)>0\); the expression approaches zero through positive values. In either case its limit is \[ \lim_{x\to-\infty}\frac{4x^2}{-8x^5}=0. \] At \(x=-2\), the original expression is \(4(4)/[-8(-32)]=16/256=1/16\), and the simplified expression is \(-1/[2(-8)]=1/16\), verifying the simplification and sign.
Common Mistake: Ignoring the Sign at Negative Infinity
A frequent error is to use the exponent difference correctly but then report the wrong signed infinite limit. For instance, \[ \frac{5x^4}{-2x^3}=-\frac52x \] has exponent difference \(1\), so its magnitude grows. But the incorrect conclusion “the limit is \(-\infty\)” treats \(x\) as positive. At negative infinity, \(x<0\), so \(-\frac52x>0\), and the correct limit is \(+\infty\).
There is a second domain error to avoid: \(\frac{x^{1/2}}{x}\) is not a real-valued expression for negative \(x\), because \(x^{1/2}\) is not real there. Subtracting exponents to get \(x^{-1/2}\) does not repair the domain problem. A real limit at negative infinity requires the original function to be real-valued for sufficiently negative inputs.
A Reliable Procedure
For a ratio of powers, use the following sequence. It keeps the domain, growth, and sign questions distinct.
At positive infinity every real power of \(x>0\) is real. At negative infinity, confirm that each power is real-valued for negative inputs.
Compute \(r=\alpha-\beta\). The magnitude behaves like a positive constant times \(|x|^r\).
If \(r<0\), the magnitude tends to zero; if \(r=0\), it stays constant; if \(r>0\), it grows without bound.
At positive infinity use the coefficient ratio. At negative infinity include the signs of the powers as well as the coefficients.
Why This Comparison Is Useful
Comparing powers extends the degree comparison used for rational functions: integer exponents are a special case of real exponents. The same reasoning also clarifies why simply saying “the numerator has the larger power” is incomplete at negative infinity. Exponent differences control size, while domain and sign determine whether the signed limit exists as a real limit and which infinite direction it takes.
Keep the distinction explicit in your work. First state the exponent difference and the resulting magnitude behavior. Then state why the powers are defined and what sign the ratio has near the direction of approach. This is especially valuable when exponents are fractional, because the domain restriction can matter as much as the growth rate.
Check Your Understanding
For each question, consider the domain, exponent difference, magnitude, and sign as appropriate.
- Evaluate \(\displaystyle\lim_{x\to+\infty}\frac{3x^{2/5}}{6x^{7/5}}\).
- Evaluate \(\displaystyle\lim_{x\to-\infty}\frac{2x^6}{-5x^4}\), and explain the sign.
- Is \(x^{3/4}\) real-valued for negative \(x\) under the usual real-power definition? Explain using the reduced denominator.
- Evaluate \(\displaystyle\lim_{x\to-\infty}\frac{x^{1/3}}{x^{4/3}}\).
- For a ratio with exponent difference \(0\), what determines its limit at positive infinity, and what additional check is needed at negative infinity?
