The role of asymptotic notation in $e^x=1+𝑥+Θ(𝑥^2)$?

I'm reading CLRS and there is the following:

When x→0, the approximation of $$e^x$$ by $$1+x$$ is quite good: $$e^x=1+𝑥+Θ(𝑥^2)$$

I suppose I understand what means this equation from math perspective and, also, there is an answer in another cs question. But I don't understand some things, so have a few questions.

1. Why do they use $$Θ$$ here and why do they use $$=$$ sign?
2. Is it possible to explain how the notation here is related to the author's conclusion that $$e^x$$ is very close to $$1 + x$$ when $$x \rightarrow 0$$?
3. Also, how is it important here that $$x$$ tends to $$0$$ rather than to $$\infty$$ as we usually use asymptotic notations?

I'm sorry if there are a lot of questions and if they are stupid, I'm just trying to master this topic.

• Note that the quoted paragraph continues: "In this equation, the asymptotic notation is used to describe the limiting behavior as $x\to 0$ rather than as $x\to \infty$". The way $\Theta$ is used here is completely different from how it is used anywhere else in the book. Jul 4 '19 at 8:26

What the equation means is that there exist constant $$A>0$$ and $$B,C$$ such that $$|x| \leq A \Longrightarrow Bx^2 \leq e^x-1-x \leq Cx^2.$$ In particular, for small $$x$$, $$e^x$$ is very close to $$1+x$$, since the error is only $$O(x^2)$$ (when $$x$$ is small, $$x^2$$ is much smaller than $$x$$).

The same estimate doesn't hold for large $$x$$ — indeed, $$e^x$$ grows faster than $$x^2$$, indeed faster than any fixed power of $$x$$. So the quoted estimate only holds for small $$x$$.

Here is a graphic example. Below is the plot of $$\frac{e^x-1-x}{x^2}$$ for $$|x| \leq 0.1$$. You can see that it is very close to $$0.5$$. This is consistent with the Taylor expansion of $$e^x$$, which is $$1 + x + x^2/2 + O(x^3)$$.

In contrast, here is the same plot for $$10 \leq x \leq 20$$. You can see that the ratio shoots up to infinity.

Credit: plots executed using Desmos.

• I think I got it. Thank you very much for your detailed answer :) Jul 5 '19 at 20:05
• @YuvalFilmus I found you answer great.. Now I felt like clarifying a few points, (1) the authors have used the asymptotic notation in a completely different meaning instead of $n\geq n_0$ they have changed the definition to $n\leq\epsilon$ where $\epsilon$ is a small positive quantity. (2) assuming (1) and as you showed $e^x-x-1$ is tight bounded by $x^2$ for very small $x$. Is this so? Jun 19 '20 at 16:02
• Right, usually big Theta is with respect to $n \to \infty$, here it is used for $x \to 0$. Jun 19 '20 at 16:48
• Taylor's theorem shows that for any $k$, we have $e^x = 1 + x + \cdots + x^k/k! + e^{\theta x} x^{k+1}/(k+1)!$, where $\theta$ is between $0$ and $x$. Jun 19 '20 at 16:50