# 向心加速度與瞬時速度垂直嗎？

(1)瞬時加速度的方向，即為瞬時速度變化的方向。
(2)當瞬時加速度即向心加速度時（例如，做等速率圓周運動的物體），
瞬時加速度與瞬時速度就是互相垂直的。

$$\large x(t)=t^{2}\tag{1}$$

\begin{align}\large v(1)\;&\large=\lim_{\Delta t\to 0}\frac{x(1+\Delta t)-x(1)}{\Delta t}\tag{2a}\\[4ex]&\large=\lim_{\Delta t\to 0}\frac{(1+\Delta t)^{2}-1^{2}}{\Delta t}\tag{2b}\\[4ex]&\large=\lim_{\Delta t\to 0}\frac{\left(2+\Delta t\right)\Delta t}{\Delta t}\tag{2c}\end{align}目前都是沒問題的。但是如果我們直接跳到底下這一步，那就要小心了。

\begin{align}\large v(1)\;&\large=\lim_{\Delta t\to 0}\left(2+\Delta t\right)\tag{3a}\label{3a}\\[4ex]&\large=2\tag{3b}\label{3b}\end{align}

\begin{align}\large\lim_{x\to 0}f(x)\;&\large=\lim_{x\to 0}\left(1+x\right)\tag{4a}\label{4a}\\[4ex]&\large=1\tag{4b}\label{4b}\end{align}

$$\large\lim_{x\to a}f(x)=L\tag{5}$$

(1*)瞬時加速度的方向，即為當 $\Delta t\to 0$ 時，瞬時速度變化的方向所趨近的方向

[1] 詳細內容請參考作者在 P87 的說明，該段落原文如下：

Suppose $f(x)$ is defined when $x$ is near the number $a$.(This means that $f$ is defined on some open interval that contains $a$, except possibly at $a$ itself.) Then we write

$$\large\lim_{x\to a }f(x)=L$$

and say “the limit of $f(x)$, as x approaches a, equals $L$” if we can make the values of $f(x)$ arbitrarily close to $L$ (as close to $L$ as we like) by taking $x$ to be sufficiently close to $a$ (on either side of $a$) but not equal to $a$.

Roughly speaking, this says that the values of $f(x)$ approach $L$ as $x$ approaches $a$. In other words, the values of $f(x)$ tend to get closer and closer to the number $L$ as $x$ gets closer and closer to the number $a$ (from either side of $a$) but $x\neq a$.

An alternative notation for

$$\large\lim_{x\to a}f(x)=L$$

is “$f(x)\to L$ as $x\to a$” which is usually read “$f(x)$ approaches $L$ as $x$ approaches $a$”.

[2] 如果同學覺得，我們要怎麼知道 f(x) 真的會在 $x$ 接近 $a$ 時，也跟著接近 $L$，而不是 $L+0.1$ 或 $L+0.00000001$ 呢？例如當 $x\to2$ 時，我該如何證明 $x^2\to2^2$ 呢？甚至，該如何證明當 $\to0$ 時，$\sin\left(x\right)/x\to1$呢？最簡單的想法是，讓我們來玩賭博遊戲吧！

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