5 Steps to a 5 AP Calculus BC 2019

Limits and Continuity 59

Example 4

Find: limx→ 2 −

[x]−x

2 −x

, where [x] is the greatest integer value ofx.

Asx→ 2 −,[x]=1. The limit of the numerator is (1−2)=−1. Asx→ 2 −,(2−x)= 0

through positive values. Thus, limx→ 2 −

[x]−x

2 −x

=−∞.

TIP • Do easy questions first. The easy ones are worth the same number of points as the hard
ones.

Limits at Infinity (asx→±∞)

If f is a function defined at every number in some interval (a,∞), then limx→∞f(x)=L

means thatLis the limit off(x)asxincreases without bound.

If fis a function defined at every number in some interval (−∞,a), then limx→−∞f(x)=L

means thatLis the limit off(x)asxdecreases without bound.

Limit Theorem

Ifnis a positive integer, then

(a) xlim→∞

1

xn

= 0

(b) xlim→−∞

1

xn

= 0

Example 1

Evaluate the limit: limx→∞

6 x− 13

2 x+ 5

.

Divide every term in the numerator and denominator by the highest power ofx(in this

case, it isx), and obtain:

xlim→∞

6 x− 13

2 x+ 5

=xlim→∞

6 −

13

x

2 +

5

x

=

xlim→∞(6)−xlim→∞

13

x

xlim→∞(2)+xlim→∞

(

5

x

)=

xlim→∞(6)−13 limx→∞

(

1

x

)

xlim→∞(2)+5 limx→∞

(

1

x

)

=

6 −13(0)

2 +5(0)

= 3.

Verify your result with a calculator. (See Figure 5.2-3.)

[–10,30] by [–5,10]

Figure 5.2-3