Why Nested Limits Need Care
A Taylor expansion can reveal a limit when both the numerator and denominator approach zero. For nested functions, however, the inner function changes the powers and coefficients in the outer expansion. A term that looks small in the outer function can matter, while a term that appears important may cancel after substitution.
The essential task is to identify the first power that remains after composing the expansions. For example, the difference between \(\log(1+x)\) and \(x\) is small near zero, but it contributes to the quadratic term of \(\sin(\log(1+x))\). Keeping only the leading approximation \(\log(1+x)\approx x\) would miss that contribution.
The method here builds on Taylor expansions for indeterminate limits: write each local expansion with a remainder, substitute the inner expansion into the outer one, and retain enough terms to determine the limit.
A Rule for Composing Local Expansions
Suppose \(u(x)\to 0\) as \(x\to 0\), and \(F\) has a local expansion about \(a\). To evaluate \(F(a+u(x))\), substitute \(u(x)\) for the increment in that expansion, then expand and collect powers of \(x\). The key is to keep terms until the first order needed by the limit is determined.
Here \(u(x)=O(x^r)\) means that \(|u(x)|\le C|x|^r\) for some constant \(C\) when \(x\) is sufficiently close to zero. To justify the remainder after substitution, write it as \(R(u(x))\), where \(R(u)=o(u^m)\). Whenever \(u(x)\ne0\), $$ \frac{R(u(x))}{x^{rm}} = \frac{R(u(x))}{u(x)^m} \left(\frac{u(x)}{x^r}\right)^m. $$ The first factor tends to zero and the second is bounded, so the product tends to zero. This proves the stated remainder order. If \(u(x)=0\) at some points, the same conclusion follows by taking \(R(0)=0\) and applying the defining remainder estimate at the other points.
This result explains why the order of the inner function matters. If \(u(x)\) begins at order \(x^r\), then \(u(x)^k\) begins at order \(x^{rk}\). Terms with \(rk\) greater than the target order cannot affect its coefficient. Terms with \(rk\) at or below the target order may matter, and their coefficients must be computed.
Worked Example: A Logarithm Inside a Sine
Worked Example: A Logarithm Inside a Sine
Evaluate $$ \lim_{x\to 0}\frac{\sin(\log(1+x))-x}{x^2}. $$ The logarithm is defined for \(x>-1\), so the expression is defined in a neighborhood of zero.
First expand the inner function: $$ \log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4). $$ Write \(u=\log(1+x)\). Since \(u=O(x)\), the sine expansion gives $$ \sin u=u-\frac{u^3}{6}+O(u^5). $$ For the terms through \(x^2\), \(u^3=O(x^3)\), so $$ \sin(\log(1+x)) =x-\frac{x^2}{2}+O(x^3). $$ Therefore, $$ \sin(\log(1+x))-x=-\frac{x^2}{2}+O(x^3). $$ Dividing by \(x^2\) gives $$ \frac{\sin(\log(1+x))-x}{x^2} =-\frac12+O(x). $$ As \(x\to0\), the \(O(x)\) term tends to zero, so the limit is \(-\frac12\).
Check: substituting the resulting quotient form \(-\frac12+O(x)\) into the limit gives \(-\frac12\), as required.
Worked Example: An Exponential of a Sine
Worked Example: An Exponential of a Sine
Evaluate $$ \lim_{x\to 0}\frac{e^{\sin x}-1-x}{x^2}. $$
Set \(u=\sin x\). The sine expansion gives $$ u=x-\frac{x^3}{6}+O(x^5), $$ so \(u=x+O(x^3)\) and \(u^2=x^2+O(x^4)\). The exponential expansion through its quadratic term is $$ e^u=1+u+\frac{u^2}{2}+O(u^3). $$ Because \(u=O(x)\), its remainder is \(O(x^3)\). Substituting the expressions for \(u\) and \(u^2\) yields $$ e^{\sin x} =1+\left(x+O(x^3)\right) +\frac12\left(x^2+O(x^4)\right)+O(x^3) =1+x+\frac{x^2}{2}+O(x^3). $$ Thus $$ \frac{e^{\sin x}-1-x}{x^2} =\frac12+O(x), $$ which tends to \(\frac12\).
Check: multiplying the quotient form \(\frac12+O(x)\) by \(x^2\) recovers \(e^{\sin x}-1-x=\frac{x^2}{2}+O(x^3)\); the constant in the limit is therefore \(\frac12\).
When Lower-Order Terms Cancel
A limit may require more terms than the first two examples. If the expression subtracts terms matching the beginning of a composite expansion, those terms cancel. One must continue until the first nonzero term is found. A reliable procedure is to set the desired denominator order before truncating: if the denominator is \(x^n\), compute the numerator through order \(x^n\), even if earlier coefficients vanish.
Worked Example: Finding the First Term After Cancellation
Evaluate $$ \lim_{x\to 0} \frac{e^{\sin x}-1-x-\frac{x^2}{2}}{x^4}. $$ The numerator has its constant, linear, and quadratic terms removed. We need the composite expansion through order four.
Write \(u=\sin x\). Through order four, $$ u=x-\frac{x^3}{6}+O(x^5). $$ Squaring gives $$ u^2 =\left(x-\frac{x^3}{6}+O(x^5)\right)^2 =x^2-\frac{x^4}{3}+O(x^6). $$ Cubing and taking the fourth power give $$ u^3=x^3+O(x^5), \qquad u^4=x^4+O(x^6). $$ The exponential expansion through the fourth power has remainder \(O(u^5)=O(x^5)\): $$ e^u=1+u+\frac{u^2}{2}+\frac{u^3}{6}+\frac{u^4}{24}+O(u^5). $$ Substituting each computed power, $$ \begin{aligned} e^{\sin x} ={}&1+\left(x-\frac{x^3}{6}\right) +\frac12\left(x^2-\frac{x^4}{3}\right) +\frac{x^3}{6}+\frac{x^4}{24}+O(x^5)\\ ={}&1+x+\frac{x^2}{2} +\left(-\frac16+\frac16\right)x^3 +\left(-\frac16+\frac1{24}\right)x^4+O(x^5)\\ ={}&1+x+\frac{x^2}{2}-\frac{x^4}{8}+O(x^5). \end{aligned} $$ The cubic coefficient cancels. The quartic coefficient is $$ -\frac16+\frac1{24}=-\frac4{24}+\frac1{24}=-\frac18. $$ Consequently, $$ \frac{e^{\sin x}-1-x-\frac{x^2}{2}}{x^4} =-\frac18+O(x), $$ and the limit is \(-\frac18\).
Check: multiplying the final quotient form by \(x^4\) gives a numerator of \(-\frac18x^4+O(x^5)\). Dividing by \(x^4\) leaves \(-\frac18+O(x)\), confirming the limit.
A Common Mistake: Replacing an Inner Function Too Early
A tempting shortcut in the first example is to use \(\log(1+x)\approx x\) and then treat \(\sin(\log(1+x))\) as if it were \(\sin x\). That replacement is adequate for finding a leading term of order \(x\), but not for a difference divided by \(x^2\). The neglected quadratic term of the logarithm is exactly the term that determines the limit.
A Practical Order-Tracking Procedure
For a nested limit, organize the computation around the denominator rather than expanding every function to an arbitrary degree. First identify the power of \(x\) in the denominator. Then determine which powers of the inner increment can contribute at or below that order. Finally, substitute and collect coefficients, paying particular attention to cancellation.
If the denominator is \(x^n\), find the numerator through order \(x^n\), unless its coefficient at that order also cancels and a higher order is needed.
Determine the leading power \(x^r\) in the inner increment. A power \(u^k\) then starts at order \(x^{rk}\).
Keep every contribution through the target order. Combine like powers before deciding which terms vanish.
After division, show that the remaining error tends to zero. This confirms that no omitted term changes the answer.
Composition is useful because it turns a nested expression into an ordinary expansion in powers of the variable approaching zero. The method also makes cancellation visible: in the exponential-of-sine example, the cubic terms from \(u\) and \(u^3/6\) cancel, so the quartic terms are needed to find the first surviving coefficient. The limit depends not merely on the leading behavior of each component, but on how their terms interact after substitution.
Check Your Understanding
For each question, track the expansion far enough to determine the requested limit.
- In \(\sin(\log(1+x))\), which term in the expansion of \(\log(1+x)\) determines the coefficient of \(x^2\)?
- If \(u(x)=O(x^2)\), what order does \(u(x)^3\) have at minimum?
- Why is \(\sin(\log(1+x))\approx\sin x\) insufficient for evaluating the first worked-example limit?
- In the third worked example, which two cubic contributions cancel?
- For a quotient with denominator \(x^5\), what is the usual target order for the numerator expansion?