# 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。

$\dpi{150}&space;x(t)=t^{2}$

$\dpi{150}&space;v(1)=\lim_{\Delta&space;t\rightarrow&space;0}\frac{x(1+\Delta&space;t)-x(1)}{\Delta&space;t}$

$\dpi{150}&space;=\lim_{\Delta&space;t\rightarrow&space;0}\frac{(1+\Delta&space;t)^{2}-1^{2}}{\Delta&space;t}=\lim_{\Delta&space;t\rightarrow&space;0}\frac{(2+\Delta&space;t)\Delta&space;t}{\Delta&space;t}$

$\dpi{150}&space;v(1)=\lim_{\Delta&space;t\rightarrow&space;0}(2+\Delta&space;t)=2$

$\dpi{150}&space;\lim_{x\rightarrow&space;0}f(x)=\lim_{x\rightarrow&space;0}(1+x)=1$

$\dpi{150}&space;\lim_{x\rightarrow&space;a}f(x)=L$

(1*)瞬時加速度的方向，即為當∆t→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

$\dpi{150}&space;\lim_{x\rightarrow&space;a}f(x)=L$

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 ≠ a.

An alternative notation for

$\dpi{150}&space;\lim_{x\rightarrow&space;a}f(x)=L$

is “f(x)→L as x→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 → 2 時，我該如何證明 x2 → 22 呢？甚至，該如何證明當 x → 0 時，sin(x)/x → 1呢？最簡單的想法是，讓我們來玩賭博遊戲吧！

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