Proving the Limit of Sin x/x Using Squeeze Theorem

[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?

 

[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.

 

[Benny]

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

 

[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

 

[gazzo]

check out the Newton quotient and the definition of the derivative

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

 

[dextercioby]

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

Daniel.

 

[Benny]

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

 

[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.

 

[dextercioby]

It's not John,but Johann.In modern French,it's l'Hôpital.

It's not certain Bernoulli did it.Wikipedia leaves a shade of doubt.

http://en.wikipedia.org/wiki/L'hopital

Daniel.

 

[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.

 

[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.

 

[dextercioby]

(...) If you define it as

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

then you can use the FTC derivative approach.(...)

.

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.

 

[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

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

 

[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...?

(...)

2. DEFINITIONS Sine and cosine are defined as the solutions to a specific differential equation with
initial conditions:
cos(x) is the unique solution to the differential equation
y"+ y= 0 subject to the initial conditions y(0)= 1, y'(0)= 0.
sin(x) is the unique solution to the differential equation
y"+ y= 0 subject to the initial conditions y(0)= 0, y'(0)= 1.
Assuming the derivatives of sine and cosine as normally developed in calculus, it is easy to
see that sine and cosine satisfy these equations. The point here is that by defining sine and cosine in
this way we can, in fact, prove all of the characteristic properties of sine and cosine from the
differential equation.
The "existence and uniqueness" theorem from the theory of differential equations tells us that
these functions do indeed exist for all x. In addition, since the differential equation is a second order
linear equation, any solution can be written as a linear combination of any two independent solutions.
Solutions satisfying these particular initial conditions: y(0)= 1, y'(0)= 0 and y(0)= 0, y'(0)= 1 are often
referred to as "fundamental" solutions. These specific values at 0 make it easy to see that the
functions are independent. It is also easy to see, simply by putting x=0 into y= C1 cos(x) + C2 sin(x)
and its derivative, that any function y satisfying y"+y= 0 subject to the conditions y(0)= A, y'(0)= B
must be y(x)= A cos(x)+ B sin(x).
3. PROPERTIES The first thing we can do is show that the usual formulas for derivatives follow
from these definitions:
Theorem 1: Derivatives: (cos(x))'= - sin(x)

(sin(x))'= cos(x)
Let u(x)= (cos(x))'. Then u'(x)= (cos(x))".
Since cosine satisfies the differential equation y"+ y=0, or
y"=-y, (cos(x))"= -(cos(x)), so u'= -(cos(x)) and
u"= -(cos(x))'=-u. Thus u satisfies u"+u=0 itself! Since
u(0)= 0 (the derivative of cos is 0 at 0) and u'(0)= -1 (cos(0)=1 and u'= -(cos(x)), we have
u(x)= 0 cos(x)+(-1) sin(x) or (cos(x))'= - sin(x).
The proof for (sin(x))' is exactly the same. Since second derivatives follow from the
differential equation- and are sine and cosine again- we immediately get derivatives of all orders and
can form the Taylor polynomial expansions for sine and cosine. That allows us to calculate
approximate values for any x. This is in
fact the way most tables for sine and cosine are constructed and the way computers and calculators
calculate values.(...)

Voilà.

Daniel.

 

[Data]

much more elegant~

 

[dextercioby]

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

Daniel.

 

[Dirac]

Limit

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

Dirac.

 

[fourier jr]

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.

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

 

[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.)

 

[arildno]

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

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