Lesson Overview
In this lesson, we'll prove that
$$\lim_{ϴ→0}\frac{sinϴ}{ϴ}=1.$$
We'll prove this result by using the squeeze theorem and basic geometry, algebra, and trigonometry. In a future lesson, we'll learn why this result is important: the reason being because knowledge that \(\lim_{ϴ→0}\frac{sinϴ}{ϴ}=1\) is required to find the derivatives of the sin and cosine functions. But we'll save that for a future lesson.
Proof of Theorem
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Alright, enough talk, let's get started with our proof. We'll start by letting \(O\) be the origin of a unit circle; next, let's draw the line \(OA\) such that the point \(A\) lies on the unit circle and has coordinates \((1,0)\). The line \(OA\) satisfying those requirements is illustrated in Figure 1. Also, we shall let the point \(B\) be any point on the unit circle in the first and fourth quadrants. We'll let the angle \(ϴ\) equal the angle \(∠BOA\). Since the point \(B\) can be anywhere along the unit circle in the first and fourth quadrants, it follows that the angle \(ϴ=∠BOA\) can vary over values in between \(\frac{-π}{2}\) and \(\frac{π}{2}\). We can represent this possible range of values of \(ϴ\) by writing the inequalities
$$\frac{-π}{2}<ϴ<\frac{π}{2}.\tag{1}$$
Figure 1: The wedge \(⪦OBA\) (colored blue) comprises a portion of the unit circle. The lengths \(AB\) and \(AC\) are the radius of the unit circle and are therefore equal to one. The heights of the triangles \(△OBA\) and \(△OCA\) can be found using basic trigonometry.
Suppose that we extend the straight line \(OB\) beyond the unit circle and that we draw a straight line which is tangent to the unit circle at the point \(A\); the point at which these two lines intersect we shall label \(C\) as illustrated in Figure 1.
As you can see visually from Figure 1, the following inequalities must be true:
$$\text{Area of triangle △OBA}≤\text{Area of wedge ⪦OBA}≤\text{Area of triangle △OCA}.\tag{2}$$
The next several steps in this proof will involve making substitutions and using algebra to rewrite Inequalities (2) in terms of \(1\), \(sinϴ/ϴ\), and \(cosϴ\); by doing so we'll be able to use the squeeze theorem to prove the statement,
$$\lim_{ϴ→0}\frac{sinϴ}{ϴ}=1.$$
We know from basic geometry that the area of any triangle is equal to \(\frac{1}{2}bh\) where \(b\) and \(h\) are the base and height of the triangle. Using this formula, we can compute that the area of the triangle \(△OBA\) must be
$$\text{Area of triangle △OBA}=\frac{1}{2}bh=\frac{1}{2}(OA)(DB).\tag{3}$$
Since the unit circle is by definition a circle whose radius equals one, we know that \(OA=1\). Substituting this result into Equations (3), we have
$$\text{Area of triangle △OBA}=\frac{1}{2}·1·DB=\frac{DB}{2}.\tag{4}$$
From basic trigonometry, we know that the sine of some angle \(ϴ\) is defined as
$$sinϴ=\frac{opposite}{hypotenuse}.$$
Using this definition, we know visually that
$$sinϴ=\frac{BD}{OB}=\frac{BD}{1}=BD.$$
Substituting this result into Equations (4), we see that
$$\text{Area of triangle △OBA}=\frac{sinϴ}{2}.\tag{5}$$
Substituting this result into Inequalities (2), we have
$$\frac{sinϴ}{2}≤\text{Area of wedge ⪦OBA}≤\text{Area of triangle △OCA}.\tag{6}$$
Let's think about how to find the area of the wedge \(⪦OBA\). We know that the area of an entire circle is given by \(πr^2\). To find the area of half a circle, we multiply the fraction \(\frac{π\text{ radians}}{2π\text{ radians}}πr^2=\frac{1}{2}πr^2\) and end up with a result that is half of the entire area \(πr^2\) of the whole circle. To find the area of a quarter of a circle, we evaluate the product \(\frac{π/2\text{ radians}}{2π \text{radians}}πr^2=\frac{1}{4}πr^2\) and end up with an area that is one quarter the area of the area \(πr^2\) of the whole circle—as expected. To find the area of the wedge \(⪦OBA\) with angle \(ϴ\), we evaluate the product
$$\frac{ϴ}{2π}πr^2=\frac{ϴ}{2}(1)^2=\frac{ϴ}{2}.$$
Thus, the area of the wedge \(⪦OBA\) is given by
$$\text{Area of ⪦OBA}=\frac{ϴ}{2}.\tag{7}$$
Substituting this result into Inequalities (6), we have
$$\frac{sinϴ}{2}≤\frac{ϴ}{2}≤\text{Area of triangle △OCA}.\tag{8}$$
The area of the triangle \(△OCA\) is given by
$$\text{Area of △OCA}=\frac{1}{2}(OA)(CA)=\frac{1}{2}(CA)\tag{9}$$
where the lengths \(OA=1\) and \(CA\) are the base and height of the triangle, respectively. We can find the height \(CA\) of the triangle \(△OCA\) using basic trigonometry. The trigonometric function \(tanϴ\) is defined as
$$tanϴ=\frac{opposite}{adjacent}.$$
Plugging \(\text{opposite }=CA\) and \(\text{adjacent }=OA=1\) into this equation, we have
$$tanϴ=\frac{CA}{1}=DB.$$
Plugging this result into Equations (9), we have
$$\text{Area of △OCA}=\frac{tanϴ}{2}.\tag{10}$$
Plugging Equation (10) into Inequalities (8), we have
$$\frac{sinϴ}{2}≤\frac{ϴ}{2}≤\frac{tanϴ}{2}.\tag{11}$$
The next part of this proof will involve making algebraic manipulations to Inequalities (11), then taking the limit as \(ϴ→0\) of all the terms in the inequalities, and then lastly using the squeeze theorem to finish the proof. Let's multiply all three terms in Inequalities (11) by 2 to get
$$|sinϴ|≤|ϴ|≤|tanϴ|.\tag{12}$$
Substituting \(tanϴ=sinϴ/cosϴ\) and dividing everything by \(sinϴ\), Inequalities (12) simplifies to
$$1≤\frac{|ϴ|}{|sinϴ|}≤\frac{1}{|cosϴ|}.\tag{13}$$
Reciprocating all three terms in Inequalities (13), we have
$$1≥\frac{|sinϴ|}{|ϴ|}≥|cosϴ|.\tag{14}$$
For any value of \(ϴ\) in between the values \(\frac{-π}{2}\) and \(\frac{π}{2}\), \(cosϴ >0\). Thus, \(|cosϴ|=cosϴ\) for these values of \(ϴ\). Also, for any values of \(ϴ\) in between \(0\) and \(\frac{π}{2}\), \(sinϴ >0\) and \(ϴ >0\). Since both \(sinϴ\) and \(ϴ\) are greater than or equal to zero when \(0<ϴ<\frac{π}{2}\), it follows that \(\frac{sinϴ}{ϴ}\) is also always greater than or equal to zero for values of \(ϴ\) within the range \(0<ϴ<\frac{π}{2}\). For values of \(ϴ\) where \(\frac{-π}{2}<ϴ<0\), \(sinϴ\) and \(ϴ\) are always negative; thus, \(sinϴ/ϴ\) must always be positive (since the quotient of two negative numbers is just a positive number) when \(\frac{-π}{2}<ϴ<0\). What we learn from putting both of these facts together is that \(sinϴ/ϴ\) is always positive for any values of \(ϴ\) in the range \(\frac{-π}{2}<ϴ<\frac{π}{2}\) such that \(ϴ≠0\) (since \(\frac{sinϴ}{ϴ}\) is undefined at \(ϴ=0\)). Thus, given these restraints for the value of \(ϴ\), it follows that \(\frac{|sinϴ|}{|ϴ|}=\frac{sinϴ}{ϴ}\). Substituting \(|cosϴ|=cosϴ\) and \(\frac{|sinϴ|}{|ϴ|}=\frac{sinϴ}{ϴ}\) into Inequalities (14), we have
$$1≥\frac{sinϴ}{ϴ}≥cosϴ.\tag{15}$$
Let's now take the limit as \(ϴ→0\) of all the terms in Inequalities (15) to get
$$\lim_{ϴ→0}1≥\lim_{ϴ→0}\frac{sinϴ}{ϴ}≥\lim_{ϴ→0}cosϴ.\tag{16}$$
The limit, \(\lim_{ϴ→0}1\), is obviously equal to 1. To find the limit, \(\lim_{ϴ→0}cosϴ\), we simply plug \(ϴ=0\) into this expression. Since \(cos0=1\), it follows that \(\lim_{ϴ→0}cosϴ=1\).
Plugging these results into Inequalities (16), we have
$$1≤\lim_{ϴ→0}\frac{sinϴ}{ϴ}≤1.\tag{17}$$
The only way for any quantity to be both greater than or equal to one and less than or equal to one is if that quantity equals one. More formally, we know from the squeeze theorem that it must be the case that
$$\lim_{ϴ→0}\frac{sinϴ}{ϴ}=1.\tag{18}$$
And we're done! In the next lesson, we'll use this result to find the derivatives of some of the trigonometric functions.
Figure 2: As you can see graphically, for values of \(ϴ\) in the range \(\frac{-π}{2}<ϴ<\frac{π}{2}\), the following inequalities are true \(1≥\frac{sinϴ}{ϴ}≥cosϴ\). Notice that as \(ϴ\) approaches zero from both the negative and positive directions, the function \(\frac{sinϴ}{ϴ}\) gets "squeezed" into the same point on the graph.
This article is licensed under a CC BY-NC-SA 4.0 license.
Sources: Khan Academy