Why the First Nonzero Term Matters
When two functions both approach zero, directly substituting into their quotient may give an indeterminate form. Taylor expansions can reveal what happens: the first nonzero term of each function describes its leading behavior near the point. Comparing those terms often determines the limit without needing the rest of either expansion.
The key information is the power of the distance from the point and the coefficient multiplying that power. If the powers match, the coefficients determine a finite limit. If the powers differ, the quotient tends to zero or grows without bound. The sign and parity of the power also matter when deciding whether a two-sided limit exists.
Leading Taylor Terms and Orders
Let \(f\) have a Taylor expansion about \(a\), and write \(h=x-a\). Suppose the first nonzero Taylor coefficient occurs at power \(m\), so that for some \(c\ne 0\),
Here \(o(h^m)\) denotes a remainder whose ratio to \(h^m\) tends to zero. Equivalently, \(f(a+h)/(h^m)\to c\). For a Taylor expansion, this asymptotic follows from Taylor’s theorem when the derivatives needed for the expansion exist near \(a\) and the remainder is of smaller order than \(h^m\).
The continuity condition matters: an asymptotic statement about \(f(a+h)\) for \(h\ne0\) alone need not determine the function’s value at \(a\). In the applications here, the functions have Taylor expansions at the point, so continuity is available.
For example, the familiar expansions at zero give \(\sin h=h+o(h)\) and \(1-\cos h=\frac12 h^2+o(h^2)\). Their leading terms have different powers: the sine vanishes to order one, while \(1-\cos h\) vanishes to order two.
The Leading-Term Quotient Rule
Suppose \(f\) and \(g\) have leading Taylor terms \(c h^m\) and \(d h^n\) at \(a\), where \(c\ne0\), \(d\ne0\), and \(h=x-a\). Assume the quotient is defined for all sufficiently small nonzero \(h\). Then
The symbol \(\sim\) here means that the ratio of the left-hand side to the right-hand side tends to \(1\). This comparison gives three main cases.
| Orders | Behavior near \(a\) |
|---|---|
| \(m>n\) | The quotient tends to \(0\). |
| \(m=n\) | The quotient tends to \(c/d\). |
| \(m<n\) | The magnitude grows without bound; parity and signs determine whether the two-sided limit is infinite or does not exist. |
When \(m<n\), the exponent \(m-n\) is negative. If it is even, \(h^{m-n}\) is positive on both sides of zero; the sign of the quotient is then the sign of \(c/d\) on both sides. If it is odd, the factor has opposite signs on the two sides, so the quotient tends to opposite infinities on the two sides. In that case the two-sided limit does not exist, even as an infinite limit.
Why the Rule Works
The conclusion follows by factoring out each leading term. From the definition of little-\(o\), the numerator can be written as its leading term times a factor approaching \(1\); the denominator has the same form. Dividing these expressions leaves the quotient of the leading terms multiplied by a factor that approaches \(1\).
Use \(f(a+h)=c h^m+o(h^m)\) and \(g(a+h)=d h^n+o(h^n)\), with both leading coefficients nonzero.
Since \(o(h^m)/h^m\to0\), write \(f(a+h)=c h^m(1+\varepsilon(h))\), where \(\varepsilon(h)\to0\). Likewise, write \(g(a+h)=d h^n(1+\delta(h))\), where \(\delta(h)\to0\).
For sufficiently small \(h\ne0\), the denominator’s factor \(1+\delta(h)\) is nonzero because it approaches \(1\). Thus the quotient equals \(\frac{c}{d}h^{m-n}\frac{1+\varepsilon(h)}{1+\delta(h)}\).
The last factor tends to \(1\). The behavior is therefore governed by \(h^{m-n}\), giving zero for a positive exponent, a constant for exponent zero, or unbounded magnitude for a negative exponent.
This argument also explains why the denominator condition is important. The quotient must be defined near the point, apart from possibly at the point itself. A denominator with nonzero leading coefficient \(d\) has the form \(d h^n(1+\delta(h))\); for sufficiently small nonzero \(h\), neither factor is zero.
Worked Examples
Worked Example: Equal Orders Give a Finite Limit
Evaluate \(\displaystyle\lim_{x\to0}\frac{\sin(3x)}{\tan(2x)}\).
Both numerator and denominator vanish at zero. From the Taylor expansions \(\sin u=u+o(u)\) and \(\tan u=u+o(u)\), substituting \(u=3x\) and \(u=2x\) gives
The leading powers are both \(x^1\); their coefficients are \(3\) and \(2\). Therefore, the quotient rule predicts the limit \(3/2\). To see the factors explicitly, write \(\sin(3x)=3x(1+\varepsilon(x))\) and \(\tan(2x)=2x(1+\delta(x))\), where \(\varepsilon(x)\to0\) and \(\delta(x)\to0\). For \(x\ne0\) sufficiently close to zero,
As a check, the first-order approximations \(3x\) and \(2x\) give the same quotient \(3x/(2x)=3/2\) whenever \(x\ne0\), and the remaining factor tends to \(1\). Hence the limit is \(\boxed{\frac32}\).
Worked Example: A Higher-Order Numerator Gives Zero
Evaluate \(\displaystyle\lim_{x\to0}\frac{x^2\cos(3x)}{(\sin x)^2}\).
The cosine has constant leading term \(1\), so
Also, \(\sin x=x+o(x)\). Squaring this expansion gives \((\sin x)^2=x^2+o(x^2)\). The numerator and denominator therefore both have order two, with leading coefficient \(1\). More explicitly, write the numerator as \(x^2(1+\varepsilon(x))\) and the denominator as \(x^2(1+\delta(x))\), with both error functions tending to zero. Then
The denominator is nonzero for sufficiently small \(x\ne0\), since \(\sin x\) has leading term \(x\). The quotient is thus defined on a punctured neighborhood of zero. The equal leading terms check the result, so the limit is \(\boxed{1}\).
Worked Example: A Higher-Order Denominator Produces Opposite Infinities
Evaluate the two-sided limit \(\displaystyle\lim_{x\to0}\frac{\sin x}{x^2}\).
The numerator has leading term \(x\), of order one, and the denominator is \(x^2\), of order two. Thus
For positive \(x\), \(1/x\to+\infty\), so the quotient tends to \(+\infty\) as \(x\to0^+\). For negative \(x\), \(1/x\to-\infty\), so the quotient tends to \(-\infty\) as \(x\to0^-\). The factor \(1+\varepsilon(x)\) tends to \(1\) and is positive sufficiently near zero, so it does not change either sign. The one-sided behaviors disagree; therefore the two-sided limit does not exist.
This checks the parity rule: the exponent difference is \(1-2=-1\), an odd negative integer, so the signs on the two sides are opposite.
Worked Example: An Even Negative Power Gives Positive Infinity
Evaluate \(\displaystyle\lim_{x\to0}\frac{(\sin x)^2}{x^4}\).
Since \(\sin x=x+o(x)\), squaring gives \((\sin x)^2=x^2+o(x^2)\). Therefore
The order of the numerator is two and the order of the denominator is four, so the power difference is \(2-4=-2\). Since \(1/x^2\to+\infty\) from both sides and the remaining factor tends to \(1\), the two-sided limit is \(+\infty\). In particular, the leading-term quotient \(x^2/x^4=1/x^2\) verifies both the growth and its positive sign. Thus the limit is \(\boxed{+\infty}\).
A Common Mistake
A frequent error is to notice that the numerator and denominator both approach zero and assume their quotient must have a finite limit. Their orders, however, may be different. For instance, a mistaken calculation for \(\sin x/x^2\) might replace \(\sin x\) by \(x\), cancel one \(x\), and then conclude that the limit is \(1\). The cancellation actually leaves \(1/x\), not \(1\):
That expression tends to opposite infinities on the two sides. The correction is to keep track of the powers after comparing leading terms: order one in the numerator minus order two in the denominator leaves power \(-1\).
Check Your Understanding
For each limit, identify the leading terms before deciding the result.
- Evaluate \(\displaystyle\lim_{x\to0}\frac{\sin(5x)}{\sin(2x)}\).
- Evaluate \(\displaystyle\lim_{x\to0}\frac{x^3\cos x}{(\sin x)^2}\).
- For \(\displaystyle\frac{x^2}{\sin x}\), what are the orders in the numerator and denominator, and what happens on each side of zero?
- If \(f(a+h)=4h^3+o(h^3)\) and \(g(a+h)=-2h^3+o(h^3)\), what is the limit of \(f(x)/g(x)\) as \(x\to a\), assuming the quotient is defined nearby?
- Why does the leading-term quotient rule require a nonzero leading coefficient for the denominator?
