24. Applications of Taylor Series

Taylor series can be used to simplify a variety of exact and approximate computations. We will look at their use in computing limits and derivatives, approximating definite integrals, summing numerical series, approximating fundamental constants and solving differential equations.

a. Limits

Compute the limit \(\displaystyle \lim_{x\to 0} \dfrac{x^2-\sin(x^2)}{x^6}\).

In the first semester of calculus you learned to compute this limit by repeatedly applying l'Hopital's Rule and simplifying: \[\begin{aligned} \lim_{x\to 0} \dfrac{x^2-\sin(x^2)}{x^6} &=\lim_{x\to 0} \dfrac{2x-2x\cos(x^2)}{6x^5} \qquad \text{apply l'Hopital's Rule} \\ &=\lim_{x\to 0} \dfrac{1-\cos(x^2)}{3x^4} \qquad \qquad \, \text{simplify} \\ &=\lim_{x\to 0} \dfrac{2x\sin(x^2)}{12x^3} \qquad \qquad \text{apply l'Hopital's Rule} \\ &=\lim_{x\to 0} \dfrac{\sin(x^2)}{6x^2} \qquad \qquad \qquad \text{simplify} \\ &=\lim_{x\to 0} \dfrac{2x\cos(x^2)}{12x} \qquad \qquad \text{apply l'Hopital's Rule} \\ &=\lim_{x\to 0} \dfrac{\cos(x^2)}{6}=\dfrac{1}{6} \qquad \qquad \text{simplify} \end{aligned}\] This is a long process.

We now recompute the limit using a series. We start with the known Maclaurin series: \[ \sin x=x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\cdots \] How do we know how many terms to keep?

and substitute \(x\to x^2\) to obtain: \[ \sin(x^2)=x^2-\dfrac{x^6}{6}+\dfrac{x^{10}}{120}-\cdots \] We now substitute this series into the limit and simplify: \[\begin{aligned} \lim_{x\to 0} \dfrac{x^2-\sin(x^2)}{x^6} &=\lim_{x\to 0} \dfrac{x^2-\left(x^2-\dfrac{x^6}{6}+\dfrac{x^{10}}{120}-\cdots\right)}{x^6} \\ &=\lim_{x\to 0} \dfrac{\dfrac{x^6}{6}-\dfrac{x^{10}}{120}+\cdots}{x^6} \\ &=\lim_{x\to 0} \left(\dfrac{1}{6}-\dfrac{x^4}{120}+\cdots\right)=\dfrac{1}{6} \end{aligned}\] Notice that the last \(+\cdots\) represents terms in which the power of \(x\) is always greater than \(4\). So the limit of these terms is zero.

Now it's your turn: