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Teaching Tip<br />

Alternate Example<br />

Finding the Limit of a Trigonometric<br />

Function<br />

x x+<br />

x<br />

Find lim 2 csc 2<br />

csc .<br />

x→0<br />

2<br />

xcsc<br />

x<br />

Solution<br />

This time, we rewrite in terms of sines:<br />

x x+<br />

x<br />

lim 2 csc 2<br />

csc<br />

x→0<br />

2<br />

xcsc<br />

x<br />

⎛ 2x<br />

1 ⎞<br />

⎜ 2 ⎟<br />

= lim sin x<br />

⎜ + sin x<br />

⎟<br />

x→0<br />

⎜<br />

x x<br />

⎟<br />

⎝<br />

2 2<br />

sin x sin x ⎠<br />

⎛ x x x ⎞<br />

= lim ⎜<br />

2 sin 2 2<br />

sin<br />

+<br />

x→0<br />

2 ⎟<br />

⎝ xsin<br />

x xsin<br />

x⎠<br />

⎛ sin x ⎞<br />

= lim +<br />

⎝<br />

⎜ 2<br />

⎠<br />

⎟ = 2 + 1 = 3<br />

x→0<br />

x<br />

Chapter objective 3: Determine where the<br />

trigonometric functions are continuous is a<br />

good review. This knowledge is a calculus<br />

prerequisite, so if the students are well<br />

prepared, you may be able to address this<br />

objective briefly to save time.<br />

AP® CaLC skill builder<br />

for example 4<br />

Finding the Limit of a Trigonometric<br />

Function<br />

<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />

122 Chapter 1 • Limits and Continuity<br />

2π π π π<br />

<br />

π<br />

2 2<br />

y<br />

1<br />

1<br />

y<br />

1<br />

1<br />

y cos x<br />

y sin x<br />

3π<br />

2<br />

2π π π π π 3π<br />

<br />

2π x<br />

2 2<br />

2<br />

Figure 45<br />

x<br />

<strong>Sullivan</strong><br />

Solution First we rewrite the expression cos θ − 1 as the product of two terms whose<br />

θ<br />

limits are known. For θ = 0,<br />

( )( )<br />

cos θ − 1 cos θ − 1 cos θ + 1<br />

=<br />

θ<br />

θ cos θ + 1<br />

= cos2 θ − 1<br />

θ(cos θ + 1) = − sin 2 ( )<br />

θ sin θ (− sin θ)<br />

↑ θ(cos θ + 1) = θ cos θ + 1<br />

sin 2 θ+ cos 2 θ = 1<br />

Now we find the limit.<br />

[( )( )] [ ][<br />

]<br />

cos θ − 1 sin θ (− sin θ)<br />

sin θ − sin θ<br />

lim = lim<br />

= lim lim<br />

θ→0 θ θ→0 θ cos θ + 1 θ→0 θ θ→0 cos θ + 1<br />

lim(− sin θ)<br />

θ→0<br />

= 1 ·<br />

lim (cos θ + 1) = 0 2 = 0<br />

■<br />

θ→0<br />

NOW WORK Problem 31 and AP® Practice Problem 8.<br />

3 Determine Where the Trigonometric Functions<br />

Are Continuous<br />

The graphs of f (x) = sin x and g(x) = cos x in Figure 45 suggest that f and g are<br />

continuous on their domains, the set of all real numbers.<br />

EXAMPLE 4 Showing f(x) = sin x Is Continuous at 0<br />

• f (0) = sin 0 = 0, so f is defined at 0.<br />

• lim f (x) = lim sin x = 0, so the limit at 0 exists.<br />

x→0 x→0<br />

• lim sin x = sin 0 = 0.<br />

x→0<br />

Since all three conditions of continuity are satisfied, f (x) = sin x is continuous at 0. ■<br />

In a similar way, we can show that g(x) = cos x is continuous at 0. That is,<br />

lim cos x = cos 0 = 1.<br />

x→0<br />

You are asked to prove the following theorem in Problem 67.<br />

THEOREM<br />

• The sine function y = sin x is continuous on its domain, all real numbers.<br />

• The cosine function y = cos x is continuous on its domain, all real numbers.<br />

Based on this theorem the following two limits can be added to the list of basic<br />

limits.<br />

lim sin x = sin c for all real numbers c<br />

x→c<br />

lim cos x = cos c for all real numbers c<br />

x→c<br />

Following are some function values for two<br />

continuous functions, f and g.<br />

x fx ( ) gx ( )<br />

1 2 −1<br />

2 1 0<br />

sin( gx ( )) + fx ( )<br />

Find lim .<br />

x→2<br />

gf (( x))<br />

Solution<br />

Here we will use the continuity of f, g, and<br />

the sine function to find the limit.<br />

sin( gx ( )) + fx ( )<br />

lim<br />

x→2<br />

gf (( x))<br />

∑ Mathematical Practices Tip<br />

MPAC 1: Reasoning with Definitions and<br />

Theorems<br />

Use the continuity of a function to find a limit.<br />

Ask students to use the definition of continuity<br />

to explain why lim cosx<br />

=−1.<br />

Have students<br />

x→π<br />

explain by using a graph how the continuity<br />

of a function f at a point x = c shows that<br />

lim f( x) = f( c).<br />

x→c<br />

sin( g(2)) + f(2)<br />

= lim<br />

x→2<br />

gf ((2))<br />

+ +<br />

= lim sin(0) 1 0 1 =<br />

− =− 1<br />

x→2<br />

g(1)<br />

1<br />

122<br />

Chapter 1 • Limits and Continuity<br />

TE_<strong>Sullivan</strong>_Chapter01_PART II.indd 5<br />

11/01/17 9:55 am

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