( )
( )
2 0sin lim
sin
x
x
x x
→ 2
2
3
2
lim
2
x
x
x
x
→
−
+
−
3 2
3 2
1
lim
3
5
2
x
x
x
x
x
x
x
→∞
+
+ +
+
+
+
2 2
lim
1
1
x→∞
x
+ −
x
−
2 2
lim
1
1
x→∞
x
+ + −
x
x
− −
x
→0 2+ + − 2− +
2 lim
2 1 3 1
x
x
x x x x
( )
0sin 3 lim
6
x
x x
→
( )
(
)
0
sin sin lim
x
x x
→
( )
( )
( )
→
+
+ + − − +
2 2
0
2sin lim
2 sin 1 sin 1
x
x x
x x x x
( )
tan x 2
lim e
x→ +π( )
=
Where y tanx is continuous?
( )
φφ
= −
2
1
Where f sin is continuous? 1
( )
( )
= − ≠− =
2
How must f 0 be determined so that f , 0, 1 is continuous at 0 ?
x x
x x
x x
( )
( )
( )
− − −
= = − = =
+ −
= =
2
0 0
0
Which of the following functions have removable singularities at the indicated points ?
2 8 1
a) f , 2, b) g , 1
2 1
1 c) h sin , 0
x x x
x x x x
x x
t t t
t
( )
= ∞Show that the equation sinx e has many solutions.x
! " # $ %
% % &
% ' & & %
# $ ' (( "
% &
% #
(
" %)
& %∞
= = ∞ ∞ × = −∞
∞
) " * +
&
∞
* # !
% ! , " * # &
(
)(
)
(
) (
)
→∞
+ − + + + −
+ − − =
+ + − + − −
= = →
+ + − + + +
&
(
)(
)
&→
− +
− = = + →
− −
-
( )
&→ =
. * !
(
+)(
−)
= − & . ! " ! #% % " ! "
# %
! $ % &
' ( / ' ( 0 ' ( 1 &
2 1
( )
( )
&→ =
) " 1 +
( )
= "( )
= ( )
= &3 ) .
-*
% &
!
4
2
2
3
2
lim
2
x
x
x
x
→
−
+
−
5
(
)(
)
2
1
2
3
2
Rewrite
1.
2
2
x
x
x
x
x
x
x
−
−
−
+ =
= −
−
−
(
)
2
2 2
3
2
Hence lim
lim
1
1.
2
x x
x
x
x
x
→ →
−
+ =
− =
−
!
4
3 2
3 2
1
lim
3
5
2
x
x
x
x
x
x
x
→∞
+
+ +
+
+
+
5
3 2 2 3
3 2
2 3
1
1
1
1
1
1.
3
5
2
3
5
2
1
x
x
x
x
x
x
x
x
x
x
x
x
x
→∞
+ +
+
+
+ +
=
→
!
4
lim
21
21
x→∞
x
+ −
x
−
5
(
2 2)(
2 2)
2 2
2 2
1 1 1 1
1 1
1 1
x x x x
x x
x x
+ − − + + −
+ − − =
+ + −
(
) (
2)
2(
) (
)
2 2 2 2
2 2 2 2 2 2
1 1 1 1 2
1 1 1 1 1 1
x x x x
x x x x x x
+ − − + − −
= = =
+ + − + + − + + −
6
2 2
2 2
2
Hence lim
1
1
lim
0.
1
1
x→∞
x
+ −
x
− =
x→∞x
+ +
x
−
=
!
4 2 2
lim
1
1
x→∞
x
+ + −
x
x
− −
x
5
(
2) (
2)
2 2 2 2
1 1 2 2
1 1 1 1
x x x x x
x x x x x x x x
+ + − − − +
= =
+ + + − − + + + − −
(
)
2 2
2 2
2 2
2 2
1 1
1 1
1 1
1 1
x x x x
x x x x
x x x x
x x x x
+ + − − − =
+ + + − −
+ + − − −
+ + + − −
2 2
2 2
2
1 1 1 1
1 1
x
x
x x x x
→∞
+
= →
+ + + − −
6
7 # % !
!
4 →
+ + − − +
2 2
0
2 lim
2 1 3 1
x
x
x x x x
5
(
)
(
) (
)
(
)
+ + + − + + + + − +
= =
+
+ + − − +
2 2 2 2
2 2 2
2 2
2 2 1 3 1 2 2 1 3 1
4
2 1 3 1
x x x x x x x x x x
x x
x x x x
(
)
(
)(
)
=
+ + − − +
+ + + − +
+ + − − + + + + − +
2 2
2 2
2 2 2 2
2
2 1 3 1
2 2 1 3 1
2 1 3 1 2 1 3 1
x
x x x x
x x x x x
x x x x x x x x
(
)
→
+ + + − +
= →
+
2 2
0
2 2 1 3 1
1
4 x
x x x x
x
6
7 # % !
!
4
( )
0
sin 3
lim
6
x
x
x
→
5
( )
( )
sin 3
1
sin 3
Rewrite
6
2
3
x
x
x
=
x
( )
0
sin
Use the fact that lim
1.
α
α
α
→
=
( )
( )
0 0
sin 3
sin 3
1
Since lim
1, we conclude that lim
.
3
6
2
x x
x
x
x
x
!
4
(
( )
)
0
sin sin lim
x
x x
→
5
( )
( )
( )
(
)
( )
00
sin
since lim 1. In the above, that fact
was applied first by substituting sin .
sin sin
Hence lim 1.
sin
x
x
x x
α
α α
α
→
→
=
=
=
( )
(
)
(
( )
)
( )
( )
0sin sin sin sin sin
1
sin x
x x x
x = x x →→
6 +
!
4
( )
( )
2 0
sin
lim
sin
xx
x
x
→
5
( )
( )
( )
( )
2 2
0 2
sin
sin
1
sin
sin
xx
x
x
x
x
=
x
x
→
→6 +
!
4
( )
( )
( )
→
+
+ + − − +
2 2
0
2 sin lim
2 sin 1 sin 1
x
x x
x x x x
5
( )
( )
( )
( )
(
)
(
( )
( )
)
( )
( )
(
)
(
( )
( )
)
+
=
+ + − − +
+ + + + − +
+ + − − + + + + − +
2 2
2 2
2 2 2 2
2 sin
2 sin 1 sin 1
2 sin 2 sin 1 sin 1
2 sin 1 sin 1 2 sin 1 sin 1
x x
x x x x
x x x x x x
x x x x x x x x
6
( )
(
)
(
( )
( )
)
( )
(
)
(
( )
)
( )
(
)
(
( )
( )
)
( )
( )
+ + + + − +
=
+ + − − +
+ + + + − +
=
− + +
2 2
2 2
2 2
2 2
2 sin 2 sin 1 sin 1
2 sin 1 sin 1
2 sin 2 sin 1 sin 1
sin 2 sin
x x x x x x
x x x x
x x x x x x
x x x x
!
4 →
( )
( )
( )
+
+ + − − +
2 2
0
2 sin lim
2 sin 1 sin 1
x
x x
x x x x
5 ' 8 (
( )
( )
( )
( )
(
)
(
( )
( )
)
( )
( )
+
+ + − − +
+ + + + − +
=
− + +
2 2
2 2
2 2
2 sin
2 sin 1 sin 1
2 sin 2sin 1 sin 1
sin 2 sin
x x
x x x x
x x x x x x
x x x x
6
( )
(
( )
( )
)
( ) ( )
( )
+ + + + − +
=
− + +
2 2
sin
1 2 2sin 1 sin 1
sin sin
sin 2 1
x
x x x x
x
x x
x x
x x
→
×
→ =
+
0
3 2 2. 2 1 x
9 '#(:# → &
"
4 tan x( )
2
lim e
x→ +π5
( )
( )
( )
2 tan
2
For , tan 0 and lim tan .
2
Hence lim e 0.
x x
x
x x x
π
π
π
π
→ +
→ +
< < < = −∞
=
4 Where the function y=tan
( )
x is continuous?5
( )
( )
( )
( )
( )
sinThe function tan is continuous whenever cos 0. cos
Hence tan is continuous at , . 2
x
y x x
x
y x x π nπ n
= = ≠
= ≠ + ∈
4
( )
21
Where the function f sin is continuous? 1
φ φ
= −
5
( )
2
1
The function f sin is continuous at all points 1
where it takes finite values. φ
φ
=
−
2 2
1 1
If 1, is not finite, and sin is undefined.
1 1
φ
φ φ
= ± − −
2 2
1 1
If 1, is finite, and sin is defined and also finite.
1 1
φ
φ φ
≠ ±
− −
2
1
Hence sin is continuous for 1.
1 φ
φ
≠ ±
−
4
( )
( )
2How must f 0 be determined so that the function
f , 0, is continuous at 0 ? 1
x x
x x x
x
−
= ≠ =
−
5
( ) ( )
( )
( )
( )
( )
(
)
0
0
0 0
2
0 0 0
Condition for continuity of a function f at a point is: lim f f . Hence f 0 must satisfy f 0 lim f .
1
Hence f 0 lim lim lim 0.
1 1
xx x
x x x
x
x x x
x x
x x
x
x x
→ →
→ → →
= =
− −
= = = =
( )
( )
00
A number for which an expression f either is undefined or
infinite is called a of the function f . The singularity is said to be , if f can be defi
singularity
removable ned in such a way that
x x
x
0
the function f becomes continous at x=x.
4
( )
( )
( )
2 0 0
0
Which of the following functions have removable singularities at the indicated points ?
2 8
a) f , 2
2 1
b) g , 1
1 1 c) h sin , 0
x x
x x
x x
x x
x
t t t
t
− −
= = −
+ −
= =
−
= =
2
6 %
6 %
7 %
( )
Show that the equation sin e hasinifinitely many solutions.
x
x =
( )
Observe that 0 e 1 for 0, and that sin 1 , .2
n x
x π nπ n
< < < + = − ∈
( )
( )
Hence f 0 for if is an odd negative number 2
and f 0 for if is an even negative number. 2
x x n n
x x n n
π π
π π < = +
> = +
4
5
sin
( )
e
xf
( )
sin
( )
e
x0.
x
=
⇔
x
=
x
−
=
;! 3 ) $
! % ! % & & & " " ! &
(
)
We conclude that every interval 2 , 2 1 , and 0, contains
2 2
a solution of the original equation. Hence there are infinitely many solutions.
n n n n
π ππ π
+ + + ∈ <
