Sine limit

**Sakai** · August 29th 2010, 04:59 AM

How do you show that $\displaystyle \lim_{x \to 0}\left(\dfrac{\sin{x}}{x}\right) = 1$ without drawing triangle areas?

**Archie Meade** · August 29th 2010, 05:06 AM

Originally Posted by Sakai

How do you show that $\displaystyle \lim_{x \to 0}\left(\dfrac{\sin{x}}{x}\right) = 1$ without drawing triangle areas?

You can use L'Hopital's Rule...

$\displaystyle\lim_{x\rightarrow\ 0}\left(\frac{Sinx}{x}\right)=\lim_{x\rightarrow\ 0}\left(\frac{\frac{d}{dx}Sinx}{\frac{d}{dx}x}\rig ht)=\lim_{x\rightarrow\ 0}\left(\frac{Cosx}{1}\right)$

Alternatively....

the area of the sector of a circle is

$\displaystyle\frac{1}{2}r^2\theta$

the area of the sector, with the segment removed is

$\displaystyle\frac{1}{2}r^2\,Sin\theta$

As the angle goes to zero, the segment area goes to zero.

$\displaystyle\frac{A_1}{A_2}=\frac{\frac{1}{2}r^2\ ,Sin\theta}{\frac{1}{2}r^2\,\theta}$

As the angle approaches zero, $A_1\rightarrow\ A_2$

Then as $\displaystyle\lim_{\theta\rightarrow\ 0}Cos\theta=1$

As $\theta\rightarrow\ 0$

$Sin\theta\ \le\theta\ \le\ Sin\theta$

**Prove It** · August 29th 2010, 07:32 AM

Originally Posted by Archie Meade

You could use L'Hopital's Rule...

$\displaystyle\lim_{x\rightarrow\ 0}\left(\frac{Sinx}{x}\right)=\lim_{x\rightarrow\ 0}\left(\frac{\frac{d}{dx}Sinx}{\frac{d}{dx}x}\rig ht)=\lim_{x\rightarrow\ 0}\left(\frac{Cosx}{1}\right)$

No you can't. Finding the derivative of $\sin{x}$ involves working out $\lim_{h \to 0}\frac{\sin{h}}{h}$ anyway. It's an entirely circular argument.

The only way that is logically correct is to squeeze the area of a sector in a unit circle by the triangles formed by $\sin{x}, \cos{x}$ and $\tan{x}, 1$ .

**Archie Meade** · August 29th 2010, 07:45 AM

Originally Posted by Prove It

No you can't. Finding the derivative of $\sin{x}$ involves working out $\lim_{h \to 0}\frac{\sin{h}}{h}$ anyway. It's an entirely circular argument.

Why not differentiate the power series expansion for Sin(x) ?

**Prove It** · August 29th 2010, 08:19 AM

Because finding the power series expansion for $\sin{x}$ ALSO involves differentiating $\sin{x}$ , which therefore would involve finding $\lim_{h \to 0}\frac{\sin{h}}{h}$ . Again, circular argument.

**Archie Meade** · August 29th 2010, 09:02 AM

Yes, that's right.

**Sakai** · August 29th 2010, 01:46 PM

Originally Posted by Prove It

Squeeze the area of a sector in a unit circle by the triangles formed by $\sin{x}, \cos{x}$ and $\tan{x}, 1$ .

I know that proof. That's why I said 'without drawing triangle areas'. I thought there might be a better way of showing it, but I guess not.

**Prove It** · August 29th 2010, 01:56 PM

The Sandwich Theorem usually provides the most elegant of limit proofs anyway...

**HallsofIvy** · August 29th 2010, 02:17 PM

I disagree with Prove It that "finding the Taylor's series for sin(x) involves differentiating sin(x)". In fact, that is one way of defining sin(x).

$\lim_{t\to 0}\frac{sin(t)}{t}$ is a very "fundamental" limit and how you do it depends in large part upon how you define sin(t).

1) A very common way of defining f(t)= sin(t) is this: On an xy- coordinate system, draw the unit circle- the circle corresponding to $x^2+ y^2= 1$ . Start at the point (1, 0) and measure (if t> 0) counterclockwise around the circumference of the circle a distance t (clockwise if t< 0). sin(t) is the y coordinate of the point you arrive at. You can proving that $\lim_{t\to 0}\frac{sin(t)}{t}= 1$ by looking at the difference between sin(t) and sin(-t) and using the area of the sector from -t to +t, sin(2t), divided by the arclength, 2t.

2) Another way of defining f(t)= sin(t) is to define $sin(t)= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}t^{2n+1}= t- \frac{1}{3!}t^3+ \frac{1}{5!}t^5+\cdot\cdot\cdot\$ .

With that definition we have, as Archie Meade suggested (though there is no reason to "differentiate" the sum), $\frac{sin(t)}{t}= \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}t^{2n}= 1- \frac{1}{3!}t^2+ \frac{1}{5!}t^4+ \cdot\cdot\cdot$ which clearly has limit 1 as x goes to 0.

3) Yet another way is to define f(t)= sin(t) to be the solution to the "intial value problem" y"+ y= 0 with initial values y(0)= 0, y'(0)= 1. From that we can prove directly that y"(0)+ y(0)= y"(0)+ 0= 0; y'''+ y'= 0 so y'''(0)+ y'(0)= y'''(0)+ 1= 0 so y'''(0)= -1; y''''+ y''= 0 so y''''(0)+ y''(0)= 0 so y''''(0)= 0, etc. and so form the Taylor's series for sin(t).

**Prove It** · August 30th 2010, 05:15 AM

Originally Posted by HallsofIvy

I disagree with Prove It that "finding the Taylor's series for sin(x) involves differentiating sin(x)". In fact, that is one way of defining sin(x).

$\lim_{t\to 0}\frac{sin(t)}{t}$ is a very "fundamental" limit and how you do it depends in large part upon how you define sin(t).

1) A very common way of defining f(t)= sin(t) is this: On an xy- coordinate system, draw the unit circle- the circle corresponding to $x^2+ y^2= 1$ . Start at the point (1, 0) and measure (if t> 0) counterclockwise around the circumference of the circle a distance t (clockwise if t< 0). sin(t) is the y coordinate of the point you arrive at. You can proving that $\lim_{t\to 0}\frac{sin(t)}{t}= 1$ by looking at the difference between sin(t) and sin(-t) and using the area of the sector from -t to +t, sin(2t), divided by the arclength, 2t.

2) Another way of defining f(t)= sin(t) is to define $sin(t)= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}t^{2n+1}= t- \frac{1}{3!}t^3+ \frac{1}{5!}t^5+\cdot\cdot\cdot\$ .

With that definition we have, as Archie Meade suggested (though there is no reason to "differentiate" the sum), $\frac{sin(t)}{t}= \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}t^{2n}= 1- \frac{1}{3!}t^2+ \frac{1}{5!}t^4+ \cdot\cdot\cdot$ which clearly has limit 1 as x goes to 0.

3) Yet another way is to define f(t)= sin(t) to be the solution to the "intial value problem" y"+ y= 0 with initial values y(0)= 0, y'(0)= 1. From that we can prove directly that y"(0)+ y(0)= y"(0)+ 0= 0; y'''+ y'= 0 so y'''(0)+ y'(0)= y'''(0)+ 1= 0 so y'''(0)= -1; y''''+ y''= 0 so y''''(0)+ y''(0)= 0 so y''''(0)= 0, etc. and so form the Taylor's series for sin(t).

I'm sorry but I totally disagree with (2) and (3).

$\sin{x}$ is NOT that polynomial by definition.

It is only written as that polynomial because you wish to find the right combination of addition, subtraction, multiplication, division and exponentiation that will enable us to evaluate $\sin{x}$ for certain values of $x$ .

If $f(x) = \sin{x}$ , the polynomial comes from

$f(x) = \sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$ .

Notice that finding these coefficients depends upon differentiating $f(x) = \sin{x}$ (infinitely many times for that matter...)

So you can not define $\sin{x}$ to be that polynomial unless you were to find the derivative of $\sin{x}$ , which again involves finding $\lim_{h \to 0}\frac{\sin{h}}{h}$ anyway.

Wherever you have to differentiate $\sin{x}$ in the definition, that means it must have been defined in some other way earlier.

Therefore, the only correct way to define $\sin{x}$ is by the unit circle definition.

**cchute** · August 30th 2010, 06:20 PM

L'Hopital's is the easiest way to do it, there is also the long way to do it (without geometric approximation). But again it's the long way and it's kind of pointless to use when you have l'hopital's. If you want to see it though look at this calculus resource under the pre-calculus section. It goes through the proof. Or you can always pop open a textbook, but hell if we all did that this forum probably wouldn't exist

**Prove It** · August 30th 2010, 11:47 PM

Read the posts I have given above.

You can NOT use L'Hospital's rule to find $\lim_{x \to 0}\frac{\sin{x}}{x}$ , because L'Hospital's Rule involves finding the derivative of $\sin{x}$ , while finding the derivative of $\sin{x}$ involves finding $\lim_{h \to 0}\frac{\sin{h}}{h}$ , the exact same limit!

Let us try to find $\frac{d}{dx}(\sin{x})$ ...

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

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

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

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

Look at that, you have to evaluate THE EXACT SAME LIMIT in order to find the derivative of $\sin{x}$ . So you can NOT use L'Hospital's Rule. It is entirely circular reasoning.

You need to have a way to evaluate this limit WITHOUT USING DERIVATIVES! That is why you have to use the Sandwich Theorem.

**Archie Meade** · August 31st 2010, 05:12 AM

However,
since the purpose of first principles is to arrive at the tangent gradient (starting with a secant),
for circular motion, we only need to take the perpendicular to the radius.

The rates of change of Sin(x) and Cos(x) can be viewed directly without reference to limits
as we consider a point rotating counterclockwise around the circumference.

If it was solely necessary to calculate the rate of change of Sin(x) using the limit of Sin(x)/x
then we would have a circular analysis.

**TheCoffeeMachine** · August 31st 2010, 07:35 PM

[tex]\displaystyle \lim_{x\to{0}}\left(\frac{\sin{x}}{x}\right) = \lim_{x\to{0}}\prod_{n=1}^{\infty}\cos\left(\frac{ x}{2^n}\right)[/Math] [LaTeX ERROR: Convert failed] .

As $x \to 0$ , $\cos\left(\frac{x}{2}\right) \to 1, \cos\left(\frac{x}{2^2}\right) \to 1, \cos\left(\frac{x}{2^3}\right) \to 1, ..., \cos\left(\frac{x}{2^n}\right) \to 1.$

Therefore [tex]\displaystyle \lim_{x\to{0}}\left(\frac{\sin{x}}{x}\right) = 1.[/Math]

Thread: Sine limit

LinkBack

Thread Tools

Display

Sine limit

Similar Math Help Forum Discussions

Limit of a function with sine

Limit, Limit Superior, and Limit Inferior of a function

Using sine theta over theta equals 1 to find a limit

help with sine

The sine limit

Search Tags