Why go to all this trouble? Because a power series turns a hard function into a polynomial, and polynomials are easy to compute with, differentiate, integrate and take limits of. This topic puts known series and error bounds to work.
Por qué funciona
Near the center, a few terms of the series already describe the function very well, and the error is controlled (by the alternating series bound or Taylor's inequality). So: a $\frac{0}{0}$ limit becomes a ratio of polynomials where the leading terms decide everything; an integral with no elementary antiderivative becomes a sum of easy power integrals; and a hard value like $\sqrt[3]{1.03}$ becomes a short arithmetic problem.
Idea clave
Replace the function by enough terms of its series. For limits, keep terms up to the power in the denominator. For integrals, integrate term by term and use the error bound to decide how many terms you need.
Ejemplo resuelto: A limit
Find $\lim_{x\to 0}\frac{\sin x - x}{x^{3}}$.
- $$\begin{aligned} &\sin x - x \\ &= \left(x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \cdots\right) - x \\ &= -\frac{x^{3}}{6} + \frac{x^{5}}{120} - \cdots. \end{aligned}$$
- Divide by $x^{3}$: $-\frac{1}{6} + \frac{x^{2}}{120} - \cdots \to -\frac{1}{6}$.
Respuesta: $-\frac{1}{6}$ (three rounds of L'Hôpital would give the same)
Ejemplo resuelto: An integral with no antiderivative
Estimate $\int_0^{1}e^{-x^{2}}\,dx$ to within $0.01$.
- $e^{-x^{2}} = 1 - x^{2} + \frac{x^{4}}{2} - \frac{x^{6}}{6} + \cdots$.
- Integrate from 0 to 1: $1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} - \cdots$.
- It alternates with shrinking terms, so stopping after $-\frac{1}{42}$ leaves an error under $\frac{1}{216} \approx 0.005$.
- Estimate: $1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} \approx 0.7429$ (true value $0.7468$).
Respuesta: $\approx 0.743$
Ejemplo resuelto: A quick value
Estimate $\ln(1.1)$ with three terms, and bound the error.
- $\ln(1 + x) \approx x - \frac{x^{2}}{2} + \frac{x^{3}}{3}$ at $x = 0.1$: $0.1 - 0.005 + 0.000333 = 0.095333$.
- Alternating with decreasing terms, so the error is less than the next term, $\frac{0.1^{4}}{4} = 0.000025$.
Respuesta: $\ln 1.1 \approx 0.09533$ (true $0.0953102$)
Error común
Keeping too few terms in a limit. In $\frac{\sin x - x}{x^{3}}$, using only $\sin x \approx x$ gives $\frac{0}{x^{3}}$, which suggests 0, the wrong answer. Keep terms until you reach the power in the denominator, here $x^{3}$. Check: after dividing, there must be a nonzero constant term (or you need more terms).