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

## 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 im 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?

## The Attempt at a Solution

So you know that if

$$|y|<\delta~\Rightarrow~|\frac{\sin(y)}{y}-1|<\varepsilon$$

You need to find

$$|x|<\delta~\Rightarrow~|\frac{\sin(2x)}{2x}-1|<\varepsilon$$

Obviously, y=2x here....

I would like to know if there's any proof of "limit of (sinx)/x as x approaches 0 is 1." ..

I would like to know if there's any proof of "limit of (sinx)/x as x approaches 0 is 1." ..

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lurflurf
Homework Helper
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.

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

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^2

Is there any proofs either than geometric reasoning? Say, something like epsilon-delta stuff? thx in advance!

Hootenanny
Staff Emeritus
Gold Member
The most straightforward method is to use the Taylor series in the vicinity of the origin:

$$\frac{\sin x}{x} = 1 - \frac{x^2}{6} +\frac{x^4}{120} + \mathcal{O}(x^6)$$

Clearly the limit is 1 as x vanishes.

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!

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?

Hootenanny
Staff Emeritus
Gold Member
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 $\sin\theta<\theta<\tan\theta$ for $\theta\in(0,\pi/2)$. 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

Thank you very much! but are there any simple proofs of x>sinx and x<tanx?

lurflurf
Homework Helper
sin(x) is chosen so that sin'(0)=1
it is just a convention