54 STEP 4. Review the Knowledge You Need to Score High
(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 limx→ 1 −[x]/=xlim→ 1 +[x], therefore, limx→ 1 [x] does not exist. (See Figure 5.1-5.)
–1
1
2
–2
y
x
–2 –1^0123
Figure 5.1-5
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 5.1-6.)
[–4,4] by [–4,4]
Figure 5.1-6
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.