Solved Problems on Limits and Continuity
3
2
2 x x + + x
1 x 3 x
2 − lim lim
3
2 x
→∞ x 2 x 3 x 5 x
2 → + + + x −
2
2
2
2
2
lim x x 1 x x
1
lim x 1 x 1 + + − − −
- − −
→∞ x x
→∞
2 x sin 3 x
( )
lim lim
→ x
6 x x 2 x + + − x 2
1 x − 2 3 x
- →
1 2
sin x ( ) lim
sin sin x x →
( ( ) ) x sin x ( ) x lim
→ x
- x 2sin x
tan x ( ) ( )
lim e
lim
- x 2 sin x
1 sin x x 1 x → + x → 2 + − − + 2 π ( ) ( )
2 Where y tan x is continuous? = ( )
) " + * ∞
∞ ∞ − ∞ &
1 ∞
Where f sin is continuous? φ =
( ) 2 ∞ −
1 φ 2 % # x − x
How must f 0 be determined so that f x , x 0, = ≠
( ) ( ) x −
1 ∞
" % & = = ∞ ∞ × ( ) = −∞ is continuous at x = 0 ?
% ∞
Which of the following functions have removable singularities at the indicated points ? 2 ! " # $ % x − 2 x − 8 x −
1 % % &
a) f x = , x = − 2, b) g x = , x =
1 ( ) ( )
- x 2 x −
1
1 % ' & & %
c) h t t sin , t = =
( ) t x # $ ' (( " % &
Show that the equation sin x = e has many solutions. ∞
( )
. ! " ! #
- & . ! ( )( − ) = − *
% % " ! " # % &
! $ −
( )( )
- − = & % & = + → → − −
2
1
- # ! % ! , " # &
- →
& ( ) = ( )
- − + +
- 1 % &
- −
( )( )
- ) "
- − − =
- −
( ) = " ( ) = ( ) = &
( ) ( + − − ) = = →
- →∞
- − + + 3 ) .
( ) ' ( / ' ( 0
= - & →
' ( 1 & !
!
3
2 x 3 x
2 −
- 2
1
- x x x
lim
4
4
lim x2 →
3
2 x
2 − x →∞ x 3 x + + + 5 x
2
2 x 1 x
2 x 3 x 2 − −
− ( )( )
5
5 Rewrite x 1.
- = = −
x 2 x
2 − −
2 x 3 x
2
1
1
1 −
2
1 Hence lim lim x
1
1. + +
- 3
- = − =
3 x x x
( )
1 x → 2 x → 2 + + x x x x
- 2
2 − 1.
3 2 = → x →∞ x 3 x 5 x
2
3
5
2
- 1
2
3 x x x
- − − −
- − − >− − − =
- − &mi>− − + + −
- − &minu
- − − −
- − − 2 2<
- −
6 7 # % ! !
- − − −
- = =
- − − + + + − − >− −
- − − =
- − + + − + + −
- = →
- − −
x x x x x →∞
- − − = =
- −
3
1
2
1
3
1
4
2
3
1
1
x x x x x x x x x x x x x x x x
( ) ( )( ) =
2
2
1
3
2
3
2
2
1
1 x
2 x x x x x x
6
4 →
2 lim
1
1
3
1 x
x x x x x
5 ( ) ( ) ( )
( )
2
2
1
2
1 Since lim 1, we conclude that lim .
1 Rewrite
x x x x x 6 7 # % !
!
4
( ) sin 36 x x x
→
5
( ) ( ) sin 3 sin 36
1
2
3 x x x x
= ( ) sin Use the fact that lim
1.
α α α
→ =
( ) ( ) sin 3 sin 3
4 x
1
1
2
3
1
2
1
3
1
1
3
3
1
x x x x x x x x x x x x x x x x x x
( ) →
2
2
1
1
1
1
1
1
1
1
2
1
1
1
( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2
1
1
x x x x x x x x x x
6
2
2
2
1
x x x x x x x x
1
→∞
!
4
2
2 lim
1
1 x x x
5 ( )( ) 2 2 2 2 2 2 2 2
1
1
1
1
1
1
1
1
2
2 Hence lim
1 1 lim 0.
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
1
1
1
2
1
x x x x x x
→∞ →∞
!
4
2
2
lim
1
1
x x x x x
→∞
5
( ) ( ) 2 2 2 2 2 21
1
2
- − − + 2 2<>− − +
- &minu
- − − + + + + − + 2 2 2 2 2 2 2 2<
- − + + + + − + = =
- − − + 2 2 2 2 2 2 2 2 2
- − + = →
- 2 2
- − − + 2 2 2 sin lim x x
- − − + 2 2 2 sin lim
- =
- − &minu
- − − +
- − + = − + + 2 2 2 2 2 2<
- − +
- − − + + + + − + 2 2 2 2 2 2 2 2<
- − + >− + =
- − &minu
- 3 2 2.
- − + = − + + 2 2 2 2 2 2 2 2
- − = = −
- ∈ <
→ → = =
2 1 x 9 '#(:# → & 7 # % !
( ) ( )
1 2 sin 2 sin 1 sin
1 2 sin 1 sin
2 sin 2 sin 1 sin
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( )
6 ( ) ( ) ( ) ( )
x x x x x x x x x x x x x x x x x x x x
x x x x x x x x x x x x x x x x x x x x
1
1 2 sin 1 sin 1 2 sin 1 sin
1 2 sin 2 sin 1 sin
2 sin 2 sin 1 sin
( ) ( ) ( )
( ) ( ) ( )
1 sin 2 sin
!
( ) ( ) ( ) ( )
→ × → =
1 x x x x x x x x x x x x
2
1 sin sin sin
1 2 2sin 1 sin
= − + + 2 2 sin
6 ( ) ( ) ( )
4 ( ) ( ) ( )
1 sin 2 sin x x x x x x x x x x x x x x x x
1 2 sin 2sin 1 sin
2 sin 2 sin 1 sin
( ) ( )
( ) ( ) ( )
( ) ( )
5 ' 8 ( ( ) ( ) ( )
→
( ) ( ) ( ) ( ) ( )
1 x x x x x x x
5 ( ) ( ) ( )
Hence lim 1. sin x
x x x x x x
sin sin sin sin sin 1 sin x
( ) ( ) ( ) ( ) ( ) ( )
= = =
α → →
α α α
x x x
1. In the above, that fact was applied first by substituting sin . sin sin
6 + !
sin since lim
( ) ( ) ( )
5 ( ) ( )
x x →
sin sin lim x
4 ( ) ( )
!
→ = →
4
( )2 sin 1 sin
2
→
4 ( ) ( ) ( )
6 + !
→ = →
x x x x x x x x
1 sin sin
sin sin
2
( )
2
( )
( ) ( )
5
( )→
x x x x
sin lim sin
2
" tan x
( ) Where the function y tan x is continuous?
4 lim e
4
= ( ) π x → +2
5
5 sin x
( ) The function y tan x is continuous whenever cos x 0.
= ( ) = ( ) ≠ cos x
( ) π For < < x , tan x < 0 and lim tan x = −∞ .
π ( ) ( ) π
π 2 x Hence y tan x is continuous at x n , n .
→ + = ( ) ≠ + π ∈
2 tan x2 ( )
Hence lim e = 0. x π → + 2
1 Where the function f = sin is continuous?
4 ( ) φ 2
4 How must f 0 be determined so that the function ( )
−
1 φ 2 x − x
f x = , x ≠ 0, is continuous at x = 0 ?
( ) 1 x −
1
5 The function f sin is continuous at all points
φ = ( ) 2
φ
1 − where it takes finite values.
5
1
1 Condition for continuity of a function f at a point x is: If φ 1, is not finite, and sin is undefined.
= ± 2 2 φ − 1 φ − 1 lim f x = f x . Hence f 0 must satisfy f 0 = lim f x . x x x ( ) ( ) ( ) ( ) ( )
→ →
1
1 If φ 1, is finite, and sin is defined and also finite.
≠ ± 2 2 2 φ − 1 φ − 1 x x −
1
x − x ( )
Hence f 0 = lim = lim = lim x = 0.
( ) x x x → → →
x
1 x
1
1 − −
Hence sin is continuous for φ ≠ ± 2 1.
φ − 1
Show that the equation sin e has inifinitely many solutions. x
< = + > = +
x = ( ) Observe that 0 e 1 for 0, and that sin 1 , .
2 n x x n n
π π
< < < + = − ∈
( ) ( ) 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 f sin e 0. x x 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 π π π
6 % 7 % ( )
2 6 %
a) f ,
( ) ( )
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 removab ned in such a way le that x x x
the function f becomes continous at . x x =
4 ( ) ( ) ( ) 2 Which of the following functions have removable singularities at the indicated points ?
2
8
2
= =
2
1
b) g ,
1
1
1
c) h sin , x x x x x x x x x t t t t
− − = = −