# Quod Erat Demonstrandum

## 2009/03/10

### Limit of tan(x)/x

Filed under: HKALE,Pure Mathematics — johnmayhk @ 9:42 下午
Tags: ,

When introducing the following ‘important’ limit to F.6B boys,

$\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$,

$\lim_{x \rightarrow \infty} \frac{\sin x}{x}$ = ?

Knowing that $\sin x$ is bounded and $\lim_{x \rightarrow \infty} \frac{1}{x} = 0$, some could give the correct answer: zero

(May be I should add one more: $\lim_{x \rightarrow 0^o} \frac{\sin x}{x}$ = ?)

Then, students tried some following up questions like:

$\lim_{x \rightarrow 0}\frac{\tan x}{x}$ = ?

It was a piece of cake to them, the answer is one.

Then an extra question turned up naturally, what is

$\lim_{x \rightarrow \infty}\frac{\tan x}{x}$ ?

Urm, up to this stage, they had no idea of l’ Hôpital rule. Even they know that rule, it don’t help much.

$\lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)}$ does not exist" does NOT imply
$\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)}$ does not exist"

Just give an old example:

$\lim_{x \rightarrow \infty}\frac{x^{-1} + \sin(x^{-1})}{x^{-1}}$ EXISTS

but

$\lim_{x \rightarrow \infty}\frac{-x^{-2} - x^{-2}\cos(x^{-1})}{-x^{-2}}$ DOES NOT.

The problem is, how to let students know (up to this stage), the way to determine whether $\lim_{x \rightarrow \infty}\frac{\tan x}{x}$ exists or not?

Taylor’s expansion? No, they had no idea!

Urm, let me put it in this way.

It is not difficult to imagine that

the graph of $y = \tan x$ intersects the graph of $y = x$ at (infinitely) many points, right?

Then $\frac{\tan x}{x} = 1$ (except at $x = 0$) at the intersections.

Besides, $\tan x = 0$ at $x = n\pi$ ($n \in \mathbb{Z}$), that means, the graph of $y = \tan x$ intersects the $x$-axis at infinitely many points.

Now, when the value of $x$ is getting large, $\frac{\tan x}{x}$ becomes zero at some moment, and sooner or later, it becomes one, and, sooner or later, it becomes zero again, and so on. Then we may FEEL that the limit of $\lim_{x \rightarrow \infty}\frac{\tan x}{x}$ does not exist.

A student, Chan, had suggested (in the lesson) rewriting the limit as

$\lim_{x \rightarrow \infty} \frac{\sin x}{x} \frac{1}{\cos x}$

but it seems that it does not help much because, at least, $\frac{1}{\cos x}$ is NOT bounded.

Welcome to share your views for better ways of teaching! Thank you for reading.

[OT]

I set up a draft of F.2 Mathematics Project in two lessons, so happy! But I still have to

1. Set up F.2 quiz (tomorrow)
2. Set up F.6 quiz (day after tomorrow)
3. Review of some questions for a moderation meeting (tomorrow)
4. Make some rearrangement about F.1 interview duty list (…)
5. Wash dishes after dinner…

TONIGHT.

Wow! Need to stop and do something meaningful now… Bye!

## 1 則迴響 »

1. Very nice explanation!

迴響 由 kong0069 — 2009/03/12 @ 9:48 上午 | 回應