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Lim θ → 0 sin(θ) / θ = 1

{$$ \frac {\sin \theta} {\theta} \tag{Fig. 1} $$}


{$$ \lim_{\theta\rightarrow0} {\frac{\sin\theta}{\theta}} = 1 $$}

I gather this is a piece of cake given the algebraic definition of sin, but we proceed here by geometry.

I am using the unit circle here, but aside from its geometric interpretation of tan, nothing here really depends upon it, so the same basic argument can be applied without it.

Since this is an argument concerning what happens when θ approaches zero, it is not necessary consider cases in which θ is greater than a right angle. A symmetric argument can be made for the case when θ approaches a straight angle, but that is not my job.

General Approach

The pinching or squeeze theorem will be used to trap

{$$ {\lim_{\theta\rightarrow0} {\frac{\sin\theta}{\theta}}} $$}

between two functions which can be shown to approach 1 as θ approaches 0.

The following Euclidian constructions are taken for granted, being established in elementary geometry:

Given the angle θ, a point O, and an arbitrary length 1

  • A circle with center 0 and radius 1 can be constructed
  • A radius of the circle (O1) can be constructed and indefinitely extended.
  • Another radius (OM) can be constructed so its angle to the radius (O1) is equal to the given angle.
  • A line perpendicular to OM can be constructed at point M and indefinitely extended.
  • MO being less than perpendicular to O1 the tangent at M will intersect the line including O1 and this point is called T.
  • A line perpendicular to O1 through the point M can be constructed. The intersection of this line and O1 is called P

By a similar method, the triangle OM'T can be constructed. Because both have two angles (θ and the right angle) equal and the lines between the angles being equal (they are radiuses of the same circle) the triangles are congruent, so the point T of one is the point T of the other.

Fig. 2

Although the line 1M is not drawn in Fig. 2, it obviously could be constructed since we know the two points 1 and M.

{$$ \text{The area of } \triangle O1M = {\frac 1 2}\sin\theta \text{ since } \overline{O1}=1$$}

Argument: The area of a triangle is ½ base x height. The line segment O1 is a radius of the unit circle and has length 1. Line segment OP was constructed perpendicular to O1, so it is the height. But it also happens to fit the geometric definition of sin θ because it is opposite θ.

{$$ \text{The arc area of } OM1 = {\frac 1 2}r^2 \theta = r^2\theta = {\frac 1 2}\theta \text{ since } r = 1 $$}

Argument: This just plugs in to the formula for the area of a sector.

{$$ \text{The area of } \triangle OMT = {\frac 1 2} \tan \theta \text{ since } \overline{OM} = 1 $$}

Argument: The area of a triangle is ½ base x height. The line segment OM is a radius of the unit circle and has length 1. Line segment MT was constructed perpendicular to OM, so it is the height. But it also happens to fit the geometric definition of tan θ.

Since triangle OM1 is wholely contained in arc sector OM1 and the sector is wholely contained in triangle OMT

{$$ {\frac 1 2}\sin\theta \leq {\frac 1 2}\theta \leq {\frac 1 2} \tan \theta $$}

Multiplying through by 2 and dividing by sin θ

But what if sin θ is a negative number? That would only be the case if θ is a negative number. If so, we should have made the geometric argument with the figure mirrored below the x-axis. But the values here would still all work out: sin θ divided by itself is still 1 where sin θ is negative; where θ is negative sin θ will be negative, so θ over sin θ is the same; and where sin θ is negative tan θ will be too (because we are only considering θ where θ is near 0, and cosine is positive when θ is near 0), leaving 1 over cos θ pristine. We should flip the inequalities here, but the function we are interested in is still in our trap.

{$$ 1 \le \frac{\theta}{\sin\theta} \le \frac{1}{\cos\theta} $$}

If you wish you can invert them all and reverse the inequalities, but if you know the limit of cos θ approaches 1 as θ approaches zero, you can quickly convince yourself its inverse does too. In either event, sin θ / θ is pinched.

{$$ \begin{align} \lim_{\theta\rightarrow0} 1 &= 1 \le \lim_{\theta\rightarrow0} {\frac{\theta}{\sin\theta}} \le \lim_{\theta\rightarrow0} {\frac{1}{\cos\theta}} = 1 \text{ or} \cr \lim_{\theta\rightarrow0} 1 &= 1 \ge \lim_{\theta\rightarrow0} {\frac{\sin\theta}{\theta}} \ge \lim_{\theta\rightarrow0} \cos\theta = 1 \end{align} $$}

We need only observe here that if anything is approaching 1, its inverse is too.

In either event, by the pinching theorem:

{$$ {\lim_{\theta\rightarrow0} {\frac{\sin\theta}{\theta}} = 1} $$}

Alternately, use lengths instead of areas to set up the pinch by observing that MT and TM' are longer than arc MM' which is in turn longer than line MPM'.


  1. FooPlot Online graphing calculator and function plotter
  2. FooPlot Online graphing calculator and function plotter



Category: Math Precalculus Limits

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This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

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August 05, 2017

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