Taylor Polynomial of $e^x$

Question

I have attached an image of section in Spivak's "Calculus", where the author explains how the difference between a function an it's Taylor's polynomial at a point $a$ is smaller than difference between $x-a$. Then the author says, that it's generalization and that for example for a $f=e^x$ the difference between first order derivative of a function and it's Taylor's polynomial is not so small as to make it higher order small function compared to difference in arguments $x-a$. He says it's easy to demonstrate it by applying L'Hospital's rule two times, but I don't understand why on denominator it's $x^2$ instead of $x$ if $a=0$. Can someone explain?

Answer 1

$$\lim_{x \to 0}\dfrac{e^x-1-x}{x}=0$$ means that, when we are very close to zero, that $e^x-1-x$ tends to zero much faster than $x$.

And $$\lim_{x \to 0}\dfrac{e^x-1-x}{x^2}=\dfrac{1}{2}$$ means that, when we are very close to zero, that $e^x-1-x$ tends to zero at approximately the same speed as $x^2$, they are "of the same order", they "look" very similar in the small neighbourhood of zero.

In other words, $\dfrac {x^2}{2}$ is a good approximation of $e^x-1-x$ when we are in a small enough neighbourhood of zero.

Answer 2

I think, the author wants to express, that the approximation using a taylor polynomial of order 1 is not as "good" as an approximation of degree 2.

More precise, using Landau-Symbols, one can show, that $f(x) - P_{k,a}(x) \in o(|x-a|^k)$ for $x \rightarrow a$.

In your explicit situation you can see, that $f(x) - P_{1, 0} \in o(x)$ for $x \rightarrow 0$, but $f(x) - P_{1, 0} \notin o(x^2)$.