Applications of Taylor — limits and infinitesimals

The most frequent use of Taylor in Year 5 is the computation of indeterminate limits: by replacing a function with its expansion, the indeterminate form is resolved in a single step.

Example — Limit via Taylor: $\lim_{x\to 0}\frac{x-\sin x}{x^3}$

Substituting the expansion $\sin x = x - x^3/6 + o(x^4)$: $\frac{x-\sin x}{x^3} = \frac{x - \bigl(x - x^3/6 + o(x^4)\bigr)}{x^3} = \frac{x^3/6 + o(x^4)}{x^3} = \frac{1}{6} + o(x) \xrightarrow{x\to 0} \boxed{\frac{1}{6}}.$ The same result as three applications of L’Hôpital, but in a single step.

Example — Order of infinitesimal

As $x\to 0$, $f(x) = 1 - \cos x$ has order of infinitesimal $2$ with respect to $x$: from the expansion $\cos x = 1 - x^2/2 + o(x^3)$ it follows that $1 - \cos x = x^2/2 + o(x^3)$, that is $1 - \cos x \sim x^2/2$.

Likewise $\sin x \sim x$ (order $1$), $e^x - 1 \sim x$ (order $1$), $\ln(1+x) \sim x$ (order $1$).

Example — Osculating parabola of $f(x) = \cos x$ at $x_0 = 0$

$f(0) = 1$, $f'(0) = 0$, $f''(0) = -1$. Taylor polynomial of order $2$: $T_2(x) = 1 + 0\cdot x + \frac{-1}{2!}x^2 = 1 - \frac{x^2}{2}.$ It is the parabola that best approximates $\cos x$ near zero. For $x = 0{,}1$ rad: $\cos(0{,}1)\approx 0{,}9950$, $T_2(0{,}1) = 1 - 0{,}005 = 0{,}995$: agreement to the first $4$ digits.

Links

Topics: Derivatives
Concepts: Order of infinitesimal · Osculating parabola · Taylor polynomial · Standard expansion
Methods: Limits via Taylor
Skills: Calculating limits · Estimating