limθ0θsinθ=?\displaystyle \lim_{\theta\to 0}\frac { \theta}{sin\theta} = ?

This note is about how to prove limθ0θsinθ \lim_{\theta\to 0}\frac { \theta}{sin\theta} using squeeze theorem.

I have created a diagram to walk us through this procedure. In the diagram there are three shapes we will be addressing. There is the triangle ABCABC (blue), the sector ABDABD (red), and the triangle AEDAED (green) which I have drawn on the unit circle.

If you are unfamiliar with the unit circle the main concept is that the unit circle has a radius of one, which is very useful for proving trigonometric identities nonetheless true in this case.

If you are unfamiliar with squeeze theorem It is the idea that if you have three quantities, quantity one, quantity two, and quantity three, where quantity one is less than or equal to quantity two, and quantity three is greater than or equal to quantity two, then at some point quantity one equals quantity three, and quantity two is also equal.


IfIf quantity 1 \leq quantity 2 \leq quantity 3

andand quantity 1 == quantity 3

thenthen quantity 1 == quantity 2 == quantity 3


In our case we will be squeezing the area of the three shapes.

area of triangle ABCABC \leq area of sector ABDABD \leq area of triangle AEDAED

Before we get into the trigonometric definition though I would like to start by pointing out that the angle θ\theta is unknown. It could be anything. If theta equals zero we have a straight line. If it is 360 degrees we have a circle.



Triangle ABCABC

Triangle ABCABC is a right angled triangle where side length ABAB is the hypoteneuse which is equal to the radius of the unit circle which is equal to one.

AB=1AB=1

SideBCBC is the oppostie side of angle θ\theta.

Side ACAC is the adjacent side to angle θ\theta

Given that:

sinθ=oppositehypoteneusesin\theta = \frac {opposite}{hypoteneuse}

sinθ=BC1sin\theta = \frac {BC}{1}

sinθ=BCsin\theta = BC

And given that:

cosθ=adjacenthypoteneusecos\theta = \frac {adjacent}{hypoteneuse}

cosθ=AC1cos\theta = \frac {AC}{1}

cosθ=ACcos\theta = AC

Then the area of the triangle is as follows:

Area=12×base×heightArea= \frac{1}{2} \times base \times height

Area=12×AC×BCArea= \frac{1}{2} \times AC \times BC

Area=12cosθsinθArea= \frac{1}{2} cos\theta sin\theta


Triangle AEDAED

Triangle AEDAED is another right angled triangle where side ADAD has a length that is equal to the radius of the unit circle which is equal to one.

AD=1AD=1

Side ADAD is the adjacent side to angle θ\theta

Side EDED is the opposite side to angle θ\theta

Given that:

tanθ=oppositeadjacent tan\theta= \frac{opposite}{adjacent}

tanθ=ED1 tan\theta= \frac{ED}{1}

tanθ=ED tan\theta= ED

And the area of triangle AEDAED is as follows:

Area=12×base×heightArea= \frac {1}{2} \times base \times height

Area=12×1×EDArea= \frac {1}{2} \times 1 \times ED

Area=12tanθArea= \frac {1}{2}tan\theta

Area=sinθ2cosθArea= \frac {sin\theta}{2cos\theta}

(tanθ=sinθcosθtan\theta= \frac{sin\theta}{cos\theta} via the quotient identity.


Sector ABDABD

Sector ABDABD is a fraction of the unit circle. So we still have our radius ADAD which is equal to one.

AD=1AD=1

The angle that makes a circle in radians is 2π2\pi so the fraction of the circle is θ2π\frac{\theta}{2\pi}.

Then the area of the sector is as follows:

Area=θ2π×π×r2Area= \frac{\theta}{2\pi} \times \pi \times r^{2}

Area=θ2π×π×(1)2Area= \frac{\theta}{2\pi} \times \pi \times (1)^{2}

Area=θ2π×π×1Area= \frac{\theta}{2\pi} \times \pi \times 1

Area=θ×π2×πArea= \frac{\theta\times\pi}{2\times\pi}

Area=θ2Area= \frac{\theta}{2}


Now we can substitute these formulas into our inequality and we will have to rearrange a few thing to find the limit we are looking for.

area of triangle ABCABC \leq area of sector ABDABD \leq area of triangle AEDAED

12cosθsinθθ2sinθ2cosθ\frac{1}{2} cos\theta sin\theta \leq \frac{\theta}{2} \leq \frac {sin\theta}{2cos\theta}

mulitply everything by 22

cosθsinθθsinθcosθ cos\theta sin\theta \leq \theta\leq \frac {sin\theta}{cos\theta}

divide everything by sinθsin \theta

cosθθsinθ1cosθ cos\theta \leq \frac{\theta}{sin\theta} \leq \frac {1}{cos\theta}

find the limit of cosθcos\theta

limθocosθθsinθ1cosθ \lim_{\theta\to o} cos\theta \leq \frac{\theta}{sin\theta} \leq \frac {1}{cos\theta}

=1θsinθ11= 1 \leq \frac{\theta}{sin\theta} \leq \frac {1}{1}

Therefore

limθ0θsinθ=1 \lim_{\theta\to 0}\frac { \theta}{sin\theta} =1

Note by Brody Acquilano
4 years, 1 month ago

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1 vote

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It's an interesting approach, going through proving that using geometry!

I was thinking of a calculus-based proof:

If we prove, using basic calculus ( Variation -derivatives ...) , that:

sinx<x<tanx;x>0\displaystyle \sin x < x < \tan x ; \forall x>0

tanx<x<sinx;x<0\displaystyle \tan x < x < \sin x ; \forall x<0,

we could also conclude the value of the limit to be 1, after calculating both handed limits.

But proving that using geometry was exciting! Cheers!!

Hasan Kassim - 4 years, 1 month ago

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Cheers.

Brody Acquilano - 4 years, 1 month ago

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It seems to me that you have merely shown that

sinθθtanθ. \sin \theta \leq \theta \leq \tan \theta.

How does that allow us to conclude what limsinθθ \lim \frac{ \sin \theta } { \theta} is?

Calvin Lin Staff - 4 years, 1 month ago

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Sorry it's finished now. Hope I did alright.

Brody Acquilano - 4 years, 1 month ago

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Right, that's the final step that's needed to draw the conclusion.

To get such "complicated" expressions to display nicely while inline, you will need to use \displaystyle, which forces the superscript / subscript, which increases out the line spacing (and hence making it uglier).

IE \displaystyle \lim_{x \rightarrow 0 } gives us limx0 \displaystyle \lim_{x \rightarrow 0 }

Calvin Lin Staff - 4 years, 1 month ago

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@Calvin Lin Thanks I will have to try that!

Brody Acquilano - 4 years, 1 month ago

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I couldn't get the 0\to 0 to go under the lim\lim that was one thing I have to fix probably.

Brody Acquilano - 4 years, 1 month ago

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By observing the given problem limit extends to 0 means the answer might be infinite(or)zero

Kutumbaka Jaswanth - 4 years, 1 month ago

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