Prove limit of (sin2x)/(2x) as x approached 0 is 1?

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In summary, the limit of (sin2x)/(2x) as x approaches 0 is 1, proven using the epsilon/delta definition of precise limit and the fact that the limit of sinx/x as x approaches 0 is 1. The proof involves finding an expression for |x|<delta that satisfies |(sin2x)/(2x) - 1|<epsilon. This can be done using the squeeze theorem or the Taylor series expansion of sinx. Additionally, the inequality sin(x)<x<tan(x) can be used to show the relationship between sinx and x.
  • #1
jessjolt
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Homework Statement


Prove limit of (sin2x)/(2x) as x approached 0 is 1. By prove I mean using the epsilon/delta definition of precise limit. You may use the fact that the limit of (sinx)/x as x approaches 0 is 1.


attempt: (where E=epsilon and d=delta)

|(sin2x)/(2x) - 1| < E if |x|<d

2(-E+1) < (sin2x)/(2x) < 2(E+1)

...now I am guessing that from here you need to isolate the x so as to get |x| is less than some expression, which solves for delta. But when I try this I keep getting that x is greater than some number, not less. Also I do not know what my professor means by being able to use the limit of sinx/x?
 
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  • #2
So you know that if

[tex]|y|<\delta~\Rightarrow~|\frac{\sin(y)}{y}-1|<\varepsilon[/tex]

You need to find

[tex]|x|<\delta~\Rightarrow~|\frac{\sin(2x)}{2x}-1|<\varepsilon[/tex]

Obviously, y=2x here...
 
  • #3
I would like to know if there's any proof of "limit of (sinx)/x as x approaches 0 is 1." ..
 
  • #4
lswtech said:
I would like to know if there's any proof of "limit of (sinx)/x as x approaches 0 is 1." ..

Sure, watch http://www.khanacademy.org/video/proof--lim--sin-x--x?playlist=Calculus [Broken]
 
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  • #5
How "limit of (sinx)/x as x approaches 0 is 1." is proven depends upon how sine has been defined. If for example sin'(0)=1 in included in the definition the result is trivial.
 
  • #6
lswtech said:
I would like to know if there's any proof of "limit of (sinx)/x as x approaches 0 is 1." ..

Think of a circle with radius r, and a differential angle dtheta that spans r*dtheta from the center on the circumference.

Sin(dtheta ) = r*dtheta / r , right ( although it seems there are 2 hypotenuses both being r, take one of them accepting the other is across the angle that is about 89.999999999... degrees )

So Sin(dtheta)*r=r*dtheta

cancel r's

to get to the eqn

Sin(dtheta)/(dtheta ) = 1

which interprets to lim x---> 0 , Sin x / x = 1
 
  • #7
Yes thanks for replies above.

But in fact I have watched somewhere else, saying that the area of circle = pi * r^2 is dependent on the result sinx/x = 1 as x->0

Say, cut the circle of radius r into n equal partitions (each with angle n/2pi in the center) and area of the circle = n*r^2 /2 *sin(n/2pi) (by area of triangle = A*B*1/2*sin(angle between AB)

Take n tends to infinity, you get pi *r^2Is there any proofs either than geometric reasoning? Say, something like epsilon-delta stuff? thanks in advance!
 
  • #8
The most straightforward method is to use the Taylor series in the vicinity of the origin:

[tex]\frac{\sin x}{x} = 1 - \frac{x^2}{6} +\frac{x^4}{120} + \mathcal{O}(x^6)[/tex]

Clearly the limit is 1 as x vanishes.
 
  • #9
Oh thank you!
But how to show that the Taylor expansion does equal to sinx without using sinx/x->1 at all? I am teaching a group of students which I want to make sure everything goes in the right path and get rid of "circular proofs"! thx!
 
  • #11
my concern is - how can we show the area of sector is 1/2*pheta?

and would there be any simple ways with epsilon-delta?
 
  • #12
lswtech said:
my concern is - how can we show the area of sector is 1/2*pheta?
As I said above, you don't need to use the geometrical setting. You can simply say that [itex]\sin\theta<\theta<\tan\theta[/itex] for [itex]\theta\in(0,\pi/2)[/itex]. This could be illustrated by plotting the functions together: http://www.wolframalpha.com/input/?i=plot+sin(x),x,tan(x)+for+x=0..pi/2
 
  • #13
Thank you very much! but are there any simple proofs of x>sinx and x<tanx?
 
  • #14
sin(x) is chosen so that sin'(0)=1
it is just a convention
 

What is the limit of (sin2x)/(2x) as x approaches 0?

The limit of (sin2x)/(2x) as x approaches 0 is equal to 1.

How do you prove the limit of (sin2x)/(2x) as x approaches 0 is 1?

The limit can be proven using the squeeze theorem or by using the fact that the limit of sinx/x as x approaches 0 is equal to 1.

What is the significance of proving the limit of (sin2x)/(2x) as x approaches 0 is 1?

Proving this limit provides a deeper understanding of the properties of trigonometric functions and their behavior near 0. It also serves as a basis for solving more complex limits involving trigonometric functions.

What are some real-life applications of this limit?

This limit has applications in fields such as physics, engineering, and astronomy. For example, it can be used to model oscillatory motion and determine the frequency of a wave.

Can this limit be generalized for other trigonometric functions?

Yes, this limit can be generalized for other trigonometric functions such as cosx/x and tanx/x. The general limit for these functions as x approaches 0 is equal to 1.

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