Why Compare Logarithms With Powers?
A logarithm increases without bound as its input increases, but it does so slowly. For example, both \(\ln x\) and \(x\) grow without bound as \(x\to\infty\). That fact alone does not tell us what happens to their quotient. The comparison we will establish is much stronger: no matter how small a positive power of \(x\) is chosen, that power eventually grows faster than \(\ln x\).
This comparison is useful when a limit contains both logarithmic and algebraic terms. It lets us determine which terms become negligible after dividing the numerator and denominator by an appropriate power of \(x\). The method is not a rule that logarithms can always be discarded: their effect depends on the powers and coefficients in the particular expression.
Proving the Basic Comparison
We prove the result with an inequality. For \(x\geq 1\), the natural logarithm has the integral representation \(\ln x=\int_1^x \frac{1}{t}\,dt\). If \(a>0\) and \(t\geq1\), then \(t^{a/2}\geq1\), so \(\frac{1}{t}\leq t^{a/2-1}\). Integrating this inequality gives a bound on \(\ln x\).
Divide by the positive quantity \(x^a\). Since \(x^{a/2}-1\leq x^{a/2}\) for \(x\geq1\), the quotient is bounded above by a simpler expression that tends to zero:
Because \(a/2>0\), the power limit \(1/x^{a/2}\to0\) as \(x\to\infty\). The lower and upper bounds therefore both approach zero, which proves the claimed limit by the squeeze theorem.
The same conclusion holds for any positive constant multiple of the logarithm: if \(c\) is fixed, then \(c\ln x/x^a\to0\). Multiplying a quantity with limit zero by a constant does not change its limit.
Positive Powers of the Logarithm
Limits sometimes involve \((\ln x)^p\), not just \(\ln x\). For every \(p>0\) and \(a>0\), the comparison still holds:
To see this, apply the proved inequality with the positive exponent \(a/p\). For \(x\geq1\), it gives \(\ln x\leq (2p/a)(x^{a/(2p)}-1)\leq (2p/a)x^{a/(2p)}\). Since both sides are nonnegative, raising them to the power \(p\) preserves the inequality. Dividing by \(x^a\) gives:
A useful equivalent form follows when the logarithm is eventually positive, as it is for \(x>1\). Taking reciprocals of positive quantities shows that \(x^a/\ln x\to+\infty\). More generally, \(x^a/(\ln x)^p\to+\infty\) for \(p>0\). This reciprocal statement is a consequence of the limit just proved; it does not mean that every expression containing a logarithm has an infinite limit.
Using the Comparison in Quotients
For a quotient containing algebraic and logarithmic terms, a useful first step is to divide every term by a common power of \(x\). Choose a power that makes the largest algebraic terms easy to identify. Then apply the logarithm-versus-power limit to any remaining logarithmic terms.
For example, after dividing by \(x^d\), a term \(x^r\) becomes \(x^{r-d}\), while a logarithmic term becomes \(\ln x/x^d\). If \(d>0\), the latter tends to zero. The exponents in the algebraic terms still matter: a term that remains constant after division can contribute to the limit, while one that tends to zero does not.
Divide numerator and denominator by a power of \(x\) that gives manageable terms.
Use power limits for algebraic terms and \(\ln x/x^a\to0\) for logarithmic terms divided by positive powers.
This step applies when the denominator has a finite, nonzero limit.
Worked Example: A Logarithm Divided by a Fractional Power
Evaluate \(\displaystyle\lim_{x\to\infty}\frac{\ln x}{x^{3/4}}\).
Use the basic theorem with \(a=3/4\). Its proof also provides an explicit bound. For \(x\geq1\),
Since \(3/8>0\), \(8/(3x^{3/8})\to0\). The expression is trapped between zero and a quantity tending to zero, so the squeeze theorem gives
Check: The expression is nonnegative for \(x\geq1\), and the upper bound tends to zero, confirming that the limit is zero.
Worked Example: A Mixed Quotient With Limit Zero
Evaluate \(\displaystyle\lim_{x\to\infty}\frac{6x^2+4\ln x}{3x^{5/2}-2x}\).
Divide every term in the numerator and denominator by \(x^{5/2}\), which is positive for \(x>0\):
The first numerator term tends to zero because \(x^{1/2}\to\infty\). The second tends to zero by the logarithm-versus-power theorem with \(a=5/2\), and \(2/x^{3/2}\to0\). Thus the numerator tends to \(0+0=0\), while the denominator tends to \(3-0=3\). Since the denominator limit is nonzero, the quotient limit is
Check: The original denominator is positive for all sufficiently large \(x\), since it equals \(x(3x^{3/2}-2)\). The numerator is also positive for \(x\geq1\). A zero limit is consistent with a positive quotient becoming arbitrarily small.
Worked Example: A Quotient With a Finite Nonzero Limit
Evaluate \(\displaystyle\lim_{x\to\infty}\frac{5x^3-2x\ln x+7}{2x^3+x^2\ln x}\).
Divide each term by \(x^3\):
By the theorem, \(\ln x/x^2\to0\) and \(\ln x/x\to0\). Also \(7/x^3\to0\). The numerator therefore tends to \(5-0+0=5\), and the denominator tends to \(2+0=2\). Since \(2\neq0\), the quotient limit is
Check: Substituting the individual term limits into the normalized quotient gives \((5-0+0)/(2+0)=5/2\), agreeing with the result.
When the Quotient Grows Without Bound
The comparison can also show that a quotient tends to positive infinity. For \(x>1\), \(\ln x>0\), and the theorem with \(a=1/2\) gives \(\ln x/x^{1/2}\to0\). Consequently \(x^{1/2}/\ln x\to+\infty\): the positive denominator is small relative to the numerator. This distinction is useful when logarithmic and algebraic terms occur in opposite parts of a quotient.
Worked Example: A Power Divided by a Logarithm
Evaluate \(\displaystyle\lim_{x\to\infty}\frac{x^{1/2}}{\ln x}\).
For \(x>1\), both \(x^{1/2}\) and \(\ln x\) are positive, so the quotient can be written as the reciprocal of a positive quantity:
The denominator in this last expression tends to zero through positive values by the theorem with \(a=1/2\). Therefore its reciprocal tends to positive infinity:
Check: The reciprocal of \(\ln x/x^{1/2}\) is exactly \(x^{1/2}/\ln x\), and the original quotient is positive for every \(x>1\). Thus the direction of the infinite limit is positive.
A Common Mistake: Treating a Logarithm Like a Power
A frequent error is to treat \(\ln x\) as if it were a degree-one algebraic term. For instance, one might incorrectly claim that \(\ln x/x\) has limit \(1\), reasoning that the numerator and denominator both “have degree one.” The logarithm is not a power \(x^1\), so polynomial degree comparisons do not apply to it.
The correction follows directly from the theorem with \(a=1\). The mistake is not merely a numerical slip: it uses a comparison rule outside its conditions. For quotients mixing logarithms and algebraic terms, normalize the expression and apply the proved growth comparison rather than assigning a polynomial degree to the logarithm.
Summary
For every \(a>0\), \(\ln x/x^a\to0\) as \(x\to\infty\). For \(p>0\), the stronger family \((\ln x)^p/x^a\to0\) also holds. Equivalently, positive powers of \(x\) eventually dominate these logarithmic terms, and reversing the quotient gives unbounded positive growth when the terms are positive.
In a mixed quotient, divide by a suitable power of \(x\), evaluate each resulting term using the appropriate limit, and take the quotient of the limits only when the denominator limit is nonzero. The logarithmic comparison identifies negligible terms; the remaining algebraic terms and coefficients determine the final result.
Check Your Understanding
Use the logarithm-versus-power comparison to answer these questions.
- For \(a>0\), what is \(\displaystyle\lim_{x\to\infty}\frac{\ln x}{x^a}\)?
- Why does the inequality \(\frac{1}{t}\leq t^{a/2-1}\) hold for \(t\geq1\) and \(a>0\)?
- What is \(\displaystyle\lim_{x\to\infty}\frac{(\ln x)^2}{x^3}\)?
- Evaluate \(\displaystyle\lim_{x\to\infty}\frac{4x^2+\ln x}{2x^2+3\ln x}\) by dividing by \(x^2\).
- Why is it incorrect to treat \(\ln x\) as a degree-one polynomial term when evaluating \(\ln x/x\)?
