5 Steps to a 5 AP Calculus BC 2019

Limits and Continuity 53

Example 1

Givenf(x)=

x^2 − 2 x− 3

x− 3

, find the limits: (a) limx→ 3 +f(x), (b) limx→ 3 −f(x), and (c) limx→ 3 f(x).

Substituting x =3 into f(x) leads to a 0 in both the numerator and denominator.

Factor f(x)as

(x−3)(x+1)

(x−3)

, which is equivalent to (x + 1) where x=/3. Thus,

(a) limx→ 3 +f(x)=xlim→ 3 +(x +1)=4, (b) limx→ 3 −f(x)=xlim→ 3 −(x +1)=4, and (c) since the

one-sided limits exist and are equal, limx→ 3 +f(x)=xlim→ 3 −f(x)=4, therefore the two-sided

limit limx→ 3 f(x) exists and limx→ 3 f(x)=4. (Note that f(x) is undefined atx=3, but the

function gets arbitrarily close to 4 asx approaches 3. Therefore the limit exists.) (See

Figure 5.1-3.)

[–8, 8] by [–6, 6]

Figure 5.1-3

Example 2

Given f(x) as illustrated in the accompanying diagram (Figure 5.1-4), find the limits:

(a) limx→ 0 − f(x), (b) limx→ 0 +f(x), and (c) limx→ 0 f(x).

[–8,8] by [–10,10]

Figure 5.1-4

(a) Asxapproaches 0 from the left, f(x) gets arbitrarily close to 0. Thus, limx→ 0 − f(x)= 0.

(b) Asxapproaches 0 from the right,f(x) gets arbitrarily close to 2. Therefore, limx→ 0 + f(x)=

2 .Note that f(0)=2.

(c) Since limx→ 0 +f(x)=xlim→ 0 −f(x), limx→ 0 f(x) does not exist.

Example 3

Given the greatest integer functionf(x)=[x], find the limits: (a) limx→ 1 +f(x), (b) limx→ 1 −f(x),

and (c) limx→ 1 f(x).

(a) Entery 1 =int(x) in your calculator. You see that asxapproaches 1 from the right, the

function stays at 1. Thus, limx→ 1 +[x]= 1 .Note that f(1) is also equal to 1.