- #1

- 23

- 0

Here is my question. How can I found that when x approaches 0, sin(x) approaches x ?

A person told me that I have to use Archimede's axioms number 27 and 28. Do someone has a demonstrationor can help me?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Ele38
- Start date

- #1

- 23

- 0

Here is my question. How can I found that when x approaches 0, sin(x) approaches x ?

A person told me that I have to use Archimede's axioms number 27 and 28. Do someone has a demonstrationor can help me?

- #2

- 22,089

- 3,296

A proof is given in the following video: http://www.khanacademy.org/video/proof--lim--sin-x--x?playlist=Calculus [Broken]

Last edited by a moderator:

- #3

disregardthat

Science Advisor

- 1,861

- 34

Here is my question. How can I found that when x approaches 0, sin(x) approaches x ?

A person told me that I have to use Archimede's axioms number 27 and 28. Do someone has a demonstrationor can help me?

Of course, you probably mean that sin(x)/x approaches 1 as x approaches 0. That sin(x) approaches x as x approaches 0 doesn't make any sense. You can prove this by [URL [Broken] rule[/url].

Last edited by a moderator:

- #4

Nabeshin

Science Advisor

- 2,205

- 16

That sin(x) approaches x as x approaches 0 doesn't make any sense.

Well hold on here, sure it does! If you taylor expand sin(x) around x=0 you get, to leading order, x.

- #5

disregardthat

Science Advisor

- 1,861

- 34

Well hold on here, sure it does!

No, "sin(x) approaches 0 as x approaches 0"

In the same way "sin(x) approaches x as x approaches 0"

That x is the first term in the Taylor series has nothing to do with this.

- #6

Nabeshin

Science Advisor

- 2,205

- 16

Would say semantics, but you're perfectly right. My physics mind was being a little lax there.

- #7

- 23

- 0

Mybe I explained in a wrong way what I wanted (sorry but I study physics no more and my english is not as googd as it should be ;) )

I would like to prove this

http://en.wikipedia.org/wiki/Small-angle_approximation

I read that I can prove the small angle aproximation with Taylor series.

[tex]sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} ...[/tex]

But when x is a very small angle I do not get how the serie converge to 0.

I would like to prove this

http://en.wikipedia.org/wiki/Small-angle_approximation

I read that I can prove the small angle aproximation with Taylor series.

[tex]sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} ...[/tex]

But when x is a very small angle I do not get how the serie converge to 0.

Last edited:

- #8

- 1,056

- 0

Of course, you probably mean that sin(x)/x approaches 1 as x approaches 0. That sin(x) approaches x as x approaches 0 doesn't make any sense. You can prove this by [URL [Broken] rule[/url].

No, that's not true. Our college physics professor put that it that way, sin(x) approaches x, as x is very small. It is a way of quickly calculating small values of sin(x). I believe he said for values under 5 degrees, transulated, of course, into radients.

All this happened before the age of calculators. Check it out, for .1 radient, 5.7 degrees, sinx = .1.

Last edited by a moderator:

- #9

uart

Science Advisor

- 2,776

- 9

Comparing the length of the arc with the length of the chord (subtended by the same angle) we can easily establish that [itex]2 \sin \theta \simeq 2 \theta[/itex] for small angles.

- #10

- 23

- 0

- #11

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

The fact that aNo, that's not true. Our college physics professor put that it that way, sin(x) approaches x, as x is very small. It is a way of quickly calculating small values of sin(x). I believe he said for values under 5 degrees, transulated, of course, into radients.

All this happened before the age of calculators. Check it out, for .1 radient, 5.7 degrees, sinx = .1.

- #12

disregardthat

Science Advisor

- 1,861

- 34

No, that's not true. Our college physics professor put that it that way, sin(x) approaches x, as x is very small. It is a way of quickly calculating small values of sin(x). I believe he said for values under 5 degrees, transulated, of course, into radients.

All this happened before the age of calculators. Check it out, for .1 radient, 5.7 degrees, sinx = .1.

Yes, as it happens sin(x) is "approximately" x around x = 0. But sin(x) --> x as x --> 0 is

Your statement as translated mathematically would require knowledge of the upper and lower bounds for (sin(x)-x) for x in [-e,e], where e is 5 degrees (in radians). By differentiating we get cos(x)-1 which has only one critical point on the interval. Thus we can assume that the largest absolute value of sin(x)-x is the maximum of |sin(e)-e|, |sin(-e)-e| and |sin(0)-0|, which is less than 0.000111. sin(x)/x --> 1 or sin(x)-x --> 0 is in fact much weaker than this, as we would have no knowledge of how close sin(x) and x are in any interval around 0.

Last edited:

- #13

- 1,056

- 0

The fact that aphysicsprofessor put it that way does NOT mean it is validmathematicsstatement! And, of course sin(.1) is NOT .1.

I did not think a physic's professor or an engineer an expert on a mathematical statement. All I was taking objection to was the statement:

Jarel: That sin(x) approaches x as x approaches 0 doesn't make any sense.

When I was working those pendulum problems in Physics years ago, we did not have calculators, and, in fact, we had not yet been taught slide rules. Numerical answers were expected, and to a Physicist approximation is what is expected in a physical problem.

Looking at the Taylor Theorem for the remainder term [tex]\frac{\tau^3}{3!}[/tex] Which in absolute value at .1 is 1.06X10^(-4). Using a calculator to two places, x-sin(x) for x=.1 is: 1.67*10^(-4).

Last edited:

Share: