5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book May 7, 2018 9:52

Limits and Continuity 89

(b) Asxapproaches 1 from the left, the function stays at 0. Therefore, limx→ 1 −[x]= 0 .Notice

that limx→ 1 −[x]=f(1).

(c) Since lim

x→ 1 −

[x]=lim

x→ 1 +

[x], therefore, lim

x→ 1

[x] does not exist. (See Figure 6.1-5.)

− 1

1

2

− 2

y

x

− 2 − 120 1 3

Figure 6.1-5

Example 4

Givenf(x)=

|x|

x

,x=0, find the limits: (a) limx→ 0 +f(x), (b) limx→ 0 −f(x), and (c) limx→ 0 f(x).

(a) From inspecting the graph, limx→ 0 +=

|x|

x

=1, (b) limx→ 0 −=

|x|

x

=−1,

and (c) since limx→ 0 +

|x|

x

=xlim→ 0 −

|x|

x

, therefore, limx→ 0 =

|x|

x

does not exist. (See Figure 6.1-6.)

[−4, 4] by [−4, 4]

Figure 6.1-6

Example 5

Iff(x)=

{

e^2 x for− 4 ≤x< 0

xex for 0 ≤x≤ 4

, find limx→ 0 f(x).

xlim→ 0 + f(x)=xlim→ 0 +xex=0 and limx→ 0 −f(x)=xlim→ 0 −e^2 x=1.

Thus, limx→ 0 f(x) does not exist.

TIP • Remember ln(e)=1 andeln3=3 sincey=lnxandy=exare inverse functions.