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Math-106, Fall2004/ MT-1 (problems& solutions )
lim 2
lim 3
lim
4 5
3 5
4 5
5
3 5
5
→∞
→∞
→∞
x x
x x
x x
x
x x
x
x
x
x
Note: We have used the following fact: =∞ x → ∞ (^) xa
lim for any a ∈ ℜ, a > 0.
We have
( )
( )
2
2
2
2
2
cos 2
cos 2
x x
x
x
x x
x
( ) 0
1 cos 2 1 2 2
2
2
− ≤ x
x x
x
x
, since 1 cos( 2 ) 1
2 − ≤ x + ≤ .By the Sandwich Theorem,
since 0
lim 2
x → −∞ x
, we deduce that
( ) 0
cos 2 lim 2
2
=
→ −∞ x
x
x
. So
( )
( )
lim 3
cos 2 lim
cos 2 lim
2
2
2
2
2
= =
− −
→−∞
→−∞
→−∞
x x
x
x
x x
x
x
x
x
cos
sin 1 lim 1
sin lim 1
tan lim 1
sin lim 1
tan 1
sin 1
lim
0
0
0
0
0
→
→
→
→
→
x x
x
x
x
x
x
x
x
x
x x
x
x x
x
x
x
x
x
Denote t x
. Then x → −∞⇒ t → 0. So, 1.
sin lim 1
sin
lim
lim sin 0
→−∞ →−∞ → t
t
x
x
x
x x x t
− − 5
1 3
1 2 5
1 3
1 2 f x 15 x 3 x 4 15 x 3 x 4
6 3
1 2 5
1 3
2 2
− − −
= ⋅ ⋅ x − ⋅ x ⋅ x + + ⋅ x − ⋅ x
6 3
1 2 5
1 2
3 2
− − −
= ⋅ x ⋅ x − ⋅ x + − ⋅ x − ⋅ x +.
lim ( ) limtan ( 1 ) tan ( 1 1 ) tan ( ) 0 0
2 2 2 2 2
1 1
→+^ →+
f x x x x
lim ( ) lim( 1 ) 1 1 0
2 2
1 1
→−^ →−
f x x x x
( ) 1 tan ( 1 1 ) tan ( ) 0 0
2 2 2 f = − = =. So f is continuous at x = 1.
( ) ( ) ( ) ( ) ( )
( )
( )
2 2
2
2
2
0
2 2
(^0 0) cos 2
sin 2 2 2
sin 2 lim
tan 2 lim
lim h h
h h h h h
h h
h
h h
h
f h f
h h h +
→+^ →+ →+
( ) ( ) lim( 2 ) 2
lim
lim 0
2
0 0
→−^ →− →−
h h
h h
h
f h f
h h h
So
( ) ( )
h
f h f
h
lim 0
→
does not exist. Hence f ′(^1 )does not exist. i.e. f is not
differentiable at x = 1.
Let u , v :Ñ → Ñ be two continuous functions that are differentiable at x ∈Ñ.
Then u^ ⋅^ v is differentiable at x^ and (^ )^ dx
dv v u dx
du u v dx
d ⋅ = ⋅ + ⋅
Proof:
( )( )
( )( ) ( )( )
h
u v x h u v x u v x dx
d
h
→ 0
lim
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
h
ux h vx h ux h vx ux h vx ux vx
h
→ 0
lim
( )
( ) ( ) ( ) ( ) ( )
→
v x h
ux h ux
h
vx h vx ux h h 0
lim
( )
( ) ( ) v ( ) x dx
dux
dx
dvx = u x ⋅ + ⋅.
[ ( ) ( )]
( ) v y v y u dx
dux vx yx
dx
d ⋅ = ⇒ ⋅ ′+ ′⋅ = ′
2 v
u v v u
v
v
u u v
v
u v y y
