Calculus Volume 1

Chapter 2 | Limits

173

Figure 2.30 The sine and tangent functions are shown as lines on the unit circle.

By dividing by sin θ in all parts of the inequality, we obtain 1< θ sin θ

< 1 cos θ .

Equivalently, we have

1> sin θ

θ >cos

θ .

sin θ

Since lim

1=1= lim

cos θ , we conclude that lim θ →0 +

θ =1.

By applying a manipulation similar to that used

θ →0 +

θ →0 +

sin θ

in demonstrating that lim θ →0 −

sin θ =0, we can show that lim θ →0 −

θ =1.

Thus,

sin θ

(2.18)

lim θ →0

θ =1.

1−cos θ θ

In Example 2.25 we use this limit to establish lim θ →0

=0.

This limit also proves useful in later chapters.

Example 2.25 Evaluating an Important Trigonometric Limit

1−cos θ θ .

Evaluate lim θ →0

Solution 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:

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