# Limit proof

1. Apr 7, 2005

### Benny

I've been looking at what I'm pretty sure is a standard result but I can't prove it. So can someone please help me out?

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$$

I think that it can be done using the squeeze theorem but I can't get to the answer. I started with $$- 1 \le \sin x \le 1$$. At this point dividing through by x gets me nowhere since the limits won't be equal. So I try to divide by cos(x) which gives: $$- \sec x \le \tan x \le \sec x$$.

The left part of the inequality approaches negative one as x approaches zero and the right part of the inequality approaches positive one as x approaches zero. I can't go any further so can someone please help me out?

2. Apr 7, 2005

### Nylex

I don't know what the squeeze theorem is, but L'Hôpital's rule will work.

lim {x->0} sin x/x = cos x/1 = 1.

3. Apr 7, 2005

### Benny

Thanks for the reply, I forgot about that. I guess I was just trying to think of an alternative way of proving it.

4. Apr 7, 2005

### josephcollins

You might also remeber to use the maclaurin's series for sinx. since sinx= x-x^3/3! + x^5/5! the x's will cancel and give 1 and the rest of the terms in x will become negligible for the given limit as x tends to zero, Joe

5. Apr 7, 2005

### gazzo

check out the newton quotient and the definition of the derivative

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - \sin 0}}{x} = \sin'(0) = \ldots$$

6. Apr 7, 2005

### dextercioby

Yes,the definition of derivative in a point is the most elegant way to do it...

Daniel.

7. Apr 7, 2005

### Benny

Thanks for the replies. I'll see what I can come up with.

8. Apr 7, 2005

### Galileo

The reason this limit is so important is because it appears when we want to compute the derivative of sin x.

From the definition:

$$\lim_{h \to 0}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h \to 0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}=\sin(x)\lim_{h \to 0}\frac{\cos(h)-1}{h}+\cos(x)\lim_{h \to 0}\frac{\sin(h)}{h}$$

The limit appears in the second term. If we want to compute this limit we cannot use l'Hospital, the series expansion or seeing it as the derivative sin x at x=0, since all these methods already use the fact that the derivative of sin(x) is cos(x).

The only rigorous way is to indeed use the squeeze theorem and find the bounds by geometric means. From geometry, you can show that:
$$\sin \theta < \theta$$
and
$$\theta < \tan \theta$$
for $\theta >0$ (but small enough, smaller than, say, $\Pi/3$. This shows that:

$$\cos \theta <\frac{\sin \theta}{\theta}<1$$

so by the squeeze theorem:

$$\lim_{\theta \to 0^+}\frac{\sin \theta}{\theta}=1$$
Since the function sinx/x is even the left side limit is also 1 and the result is proved.

From this result we can also show that the limit of (cos(x)-1)/x goes to 0, so we have proved that the derivative of sin(x) is cos(x).

l'Hospital's rule (which he didn't invent himself, but John Bernoulli did) is often used in indeterminate forms like a derivative. This method is fraud for the very simple reason that you wish to compute a derivative and differentiate the function to get it.
This is one of the reasons why I`m not a fan of l'Hospital.

Last edited: Apr 7, 2005
9. Apr 7, 2005

### dextercioby

10. Apr 7, 2005

### Data

Well, it depends how you define the $\sin$ function. If you define it by its MacLaurin series, then you can use that approach. If you define it as

$$\int_0^x \cos t \ dt$$

then you can use the FTC derivative approach.

If you define it geometrically (and since geometry was almost certainly the motivation for thinking about it at all, this probably does make the most sense), then you must approach it geometrically, as you say.

Last edited: Apr 7, 2005
11. Apr 7, 2005

### Galileo

John and Johann Bernoulli are one and the same person. And I really rather write an extra 's' then use some obscure 'alt+something' code.
Besides, what's in a name.

12. Apr 7, 2005

### dextercioby

Could you elaborate?

Oh,and yes,the method of MacLaurin series is very good,just as long as u don't mention the words "MacLaurin series"...

Daniel.

13. Apr 7, 2005

### Data

If you define

$$\int_0^x \cos t \ dt = \sin x$$

then by the FTC (fundamental theorem of calculus, if that's what you wanted me to elaborate on), you get

$$\frac{d}{dx}\sin x = \cos x$$

so you can use the argument that

$$\lim_{x \rightarrow 0} \frac{\sin x}{x} = \left(\frac{d}{dx}\sin x\right) \biggr |_{x = 0} = \cos 0 = 1.$$

14. Apr 7, 2005

### dextercioby

There's another flawless method.

Define:

$$\left\frac{df}{dx}\right|_{x_{0}} =:\lim_{x\rightarrow x_{0}}\frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}$$

Apply to the case

$$f(x)=\sin x$$

and get the much desired limit

$$\lim_{x\rightarrow 0}\frac{\sin x}{x} =\sin'(x)_{x=0}$$

Okay?Who's "sinus",why is $\sin 0=0$ (used above in the limit) and what is its derivative at x=0...?

Voilà.

Daniel.

15. Apr 7, 2005

### Data

much more elegant~

16. Apr 7, 2005

### dextercioby

The really nice part in the proof (axiomatical construction) is the periodicity proof...

Daniel.

17. Apr 10, 2005

### Dirac

Limit

$$f(x)=\lim_{x\rightarrow 0}(\frac{\sin x}{x})\\ =\lim_{x\rightarrow 0}\left( \frac{x-\frac{1}{6}x^{3}+\frac{1}{120}x^{5}...}{x}\right) \\ =\lim_{x\rightarrow 0}\left( 1-\frac{1}{6}x+\frac{1}{120}x^{4}\right)\\ x=0->\lim_{x\rightarrow 0}(\frac{\sin x}{x})=1$$

Dirac.

18. Apr 10, 2005

### fourier jr

or johannes... johannes is the catholic version of john (or the other way around). james & jakob bernoulli is the same person also, & for the same reason

19. Apr 10, 2005

### dextercioby

Nope,it's in the language."Jean" it's typically French,"Johann" is typically German,"Giovanni" is typically Italian and last,but not least,come "John" and "Juan"...

Daniel.

P.S.And "Ion" (Rom.) and "Ivan" (Rus.)

20. Apr 10, 2005

### arildno

PPS: "Jon, Jens, Johan" (Norw.)

PPPS: I thought "james" was related to the Gaelic forms "seamus, sean"?