The Limit Laws: Learn It 4

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The Limit Laws: Learn It 4

The Squeeze Theorem Cont.

We now use the squeeze theorem to tackle several very important limits. Although this discussion is somewhat lengthy, these limits prove invaluable for the development of the material in both the next section and the next module. The first of these limits is $\underset{\theta \to 0}{\lim} \sin \theta$ .

Evaluating the Limit of Sine as Theta Approaches Zero

Consider the behavior of $\sin(\theta)$ as $\theta$ approaches zero. On the unit circle, $\sin(\theta)$ corresponds to the $y$ -coordinate, which also represents the arc’s height for a given angle, $\theta$ .

As $\theta$ gets closer to zero, particularly for $0 < \theta < \frac{\pi}{2}$ , $\sin(\theta)$ becomes smaller and approaches the angle’s measure itself, meaning $\sin(\theta)$ is squeezed between $0$ and $\theta$ .

A diagram of the unit circle in the x,y plane – it is a circle with radius 1 and center at the origin. A specific point (cos(theta), sin(theta)) is labeled in quadrant 1 on the edge of the circle. This point is one vertex of a right triangle inside the circle, with other vertices at the origin and (cos(theta), 0). As such, the lengths of the sides are cos(theta) for the base and sin(theta) for the height, where theta is the angle created by the hypotenuse and base. The radian measure of angle theta is the length of the arc it subtends on the unit circle. The diagram shows that for 0 < theta < pi/2, 0 < sin(theta) < theta. — Figure 7. The sine function is shown as a line on the unit circle.

First, consider the established inequalities for $\sin(\theta)$ when $\theta$ is between $0$ and $\frac{\pi}{2}$ :

$0 < \theta < \frac{\pi}{2} \Longrightarrow 0 < \sin \theta < \theta$

Now, as $\theta$ approaches zero from the positive direction, we know that $\sin(\theta)$ also approaches zero because it is sandwiched between $0$ and $\theta$ .

Mathematically, this can be expressed as:

$\underset{\theta \to 0^+}{\lim}0=0 \text{ and }\underset{\theta \to 0^+}{\lim} \theta =0$ ,

which, according to the Squeeze Theorem, compels $\sin(\theta)$ to satisfy:

$\underset{\theta \to 0^+}{\lim} \sin \theta =0$ .

The same principle applies when approaching zero from the negative side, where $\sin(\theta)$ is negative but greater than $-\theta$ :

$\frac{-\pi}{2} < \theta < 0 \Longrightarrow-\theta < \sin \theta < 0$

Here too, as

$\theta$ approaches zero,

$\sin(\theta)$ is “squeezed” to zero:

$\underset{\theta \to 0^-}{\lim}0=0 \text{ and }\underset{\theta \to 0^-}{\lim} (-\theta) =0$ ,

leading to the conclusion that:

$\underset{\theta \to 0^-}{\lim} \sin \theta =0$ .

Therefore, we can definitively state that the limit of

$\sin(\theta)$ as

$\theta$ approaches zero from either direction is

$0$ .

the limit of $\sin(\theta)$

$\underset{\theta \to 0}{\lim} \sin \theta =0$

Evaluating the Limit of Cosine as Theta Approaches Zero

To evaluate the limit of $\cos(\theta)$ as $\theta$ approaches zero, we rely on the fundamental Pythagorean identity which states that for any angle $\theta$ , the square of the cosine of $\theta$ plus the square of the sine of $\theta$ equals one:

$\cos^2(\theta)+\sin^2(\theta)=1$

Rearranging this identity, we can isolate $\cos(\theta)$ :

$\cos(\theta)=\sqrt{1−\sin^2(\theta)}$

Since the sine function is bounded between $-1$ and $1$ for all $\theta$ , and as $\theta$ approaches zero, $\sin(\theta)$ also approaches zero, we can substitute this limit into our identity:

$\underset{\theta \to 0}{\lim} \cos \theta=\underset{\theta \to 0}{\lim} \sqrt{1−\sin^2(\theta)}$

Given that $\underset{\theta \to 0}{\lim} \sin \theta =0$ , we then have:

$\underset{\theta \to 0}{\lim} \sqrt{1−\sin^2(\theta)} =\sqrt{1-0^2}=1$

Thus, we confirm that the limit of $\cos(\theta)$ as $\theta$ approaches zero is $1$ .

the limit of $\cos(\theta)$

$\underset{\theta \to 0}{\lim} \cos \theta =1$

Exploring the Limit of Sine Theta Over Theta

A pivotal limit in calculus, particularly relevant in the study of derivatives and integrals of trigonometric functions, is

$\underset{\theta \to 0}{\lim}\frac{\sin \theta}{\theta}$ .

To understand this limit, we look to the unit circle, where the sine and tangent functions provide geometric insights into this foundational limit.

The same diagram as the previous one. However, the triangle is expanded. The base is now from the origin to (1,0). The height goes from (1,0) to (1, tan(theta)). The hypotenuse goes from the origin to (1, tan(theta)). As such, the height is now tan(theta). It shows that for 0 < theta < pi/2, sin(theta) < theta < tan(theta). — Figure 8. The sine and tangent functions are shown as lines on the unit circle.

Analyze the behavior of $\sin(\theta)$ and $\tan(\theta)$ within the first quadrant of the unit circle, specifically for angles $\theta$ where $0 < \theta < \frac{\pi}{2}$ .

In this range, it’s clear from the geometric representation that $\sin(\theta)$ is always less than the length of the tangent line segment from the point on the circle to the $x$ -axis, which is $\tan(\theta)$ . Consequently, we have the inequality:

$0< \sin \theta < \tan \theta$

By dividing each term in the inequality by $\sin \theta$ , we are led to:

$1 < \dfrac{\theta}{\sin \theta} < \dfrac{1}{\cos \theta}$

With the reciprocal, this inequality can be restated as:

$1 > \dfrac{\sin \theta}{\theta} > \cos \theta$

As $\theta$ approaches zero, $\cos(\theta)$ approaches $1$ . Therefore, $\sin(\theta)$ is squeezed between $\cos(\theta)$ and $1$ .

Since $\cos(\theta)$ also approaches $1$ as $\theta$ approaches zero, the Squeeze Theorem can be applied to conclude that:

$\underset{\theta \to 0}{\lim}\dfrac{\sin \theta}{\theta}=1$

the limit of $\dfrac{\sin \theta}{\theta}$

$\underset{\theta \to 0}{\lim}\dfrac{\sin \theta}{\theta}=1$

Evaluating the Limit of $\dfrac{1- \cos \theta}{\theta}$

As we build upon the understanding of limits involving trigonometric functions, the next step is to apply the Squeeze Theorem to evaluate limits that are not immediately obvious.

In the example below, we use the limit of $\frac{\sin {\theta}}{\theta}$ to establish $\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta}=0$ . This limit also proves useful in later modules.

Evaluate $\underset{\theta \to 0}{\lim}\dfrac{1- \cos \theta}{\theta}$

In the first step, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine in the numerator to a sine:

$\begin{array}{cc} \underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta}& =\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta} \cdot \frac{1+ \cos \theta}{1+ \cos \theta} \\ & =\underset{\theta \to 0}{\lim}\frac{1-\cos^2 \theta}{\theta(1+ \cos \theta)} \\ & =\underset{\theta \to 0}{\lim}\frac{\sin^2 \theta}{\theta(1+ \cos \theta)} \\ & =\underset{\theta \to 0}{\lim}\frac{\sin \theta}{\theta} \cdot \frac{\sin \theta}{1+ \cos \theta} \\ & =1 \cdot \frac{0}{2}=0 \end{array}$

Therefore,

$\underset{\theta \to 0}{\lim}\dfrac{1- \cos \theta}{\theta}=0$

the limit of $\dfrac{1- \cos \theta}{\theta}$

$\underset{\theta \to 0}{\lim}\dfrac{1- \cos \theta}{\theta}=0$

The Limit Laws: Learn It 4

The Squeeze Theorem Cont.

Evaluating the Limit of Sine as Theta Approaches Zero

the limit of sin(θ)\sin(\theta)

Evaluating the Limit of Cosine as Theta Approaches Zero

the limit of cos(θ)\cos(\theta)

Exploring the Limit of Sine Theta Over Theta

the limit of sinθθ\dfrac{\sin \theta}{\theta}

Evaluating the Limit of 1−cosθθ\dfrac{1- \cos \theta}{\theta}

the limit of 1−cosθθ\dfrac{1- \cos \theta}{\theta}

the limit of $\sin(\theta)$

the limit of $\cos(\theta)$

the limit of $\dfrac{\sin \theta}{\theta}$

Evaluating the Limit of $\dfrac{1- \cos \theta}{\theta}$

the limit of $\dfrac{1- \cos \theta}{\theta}$