Post on 11-Jan-2016
If is measured in radian
Then :
If is measured in radian
Then :
and:
< -1-cos <
< -sin <
From the theorem , and by sandwich theorem ,
We find that
a < < - )
sin
0 0 0As 0
lim0
sin =0
And b ) < -1-cos < so
0 0 0As 0
lim0 1-cos =0
lim0 Cos = 1
ExampleFind the limit if it exists:
0
sinlim
Example
g()=1
h()=cos
sin( )f
Example
0lim1 1
0
lim cos 1
&
therefore…
0
sinlim 1
sincos 1
If we graph , it appears thatsin x
yx
0
sinlim 1x
x
x
Find the following limits:
a)x
x3)3sin(lim
0x
Put = 3x so as x 0 , = 3x 0
1)sin(
=lim0xx
x3)3sin(= lim
0x
Solution
xx
5)3sin(b) lim
0x
= 53
xx
5)3sin(
=lim0x 5
3x
x3)3sin(
=lim0x 5
3x
x3)3sin(lim
0x
Solution
Put = sin x so as x 0 ,
= sin x 0
=
00
Find xx
sin)sin(sinlim
0x
xx
sin)sin(sinlim
0x
= = 1x
xsin
)sin(sinlim0x
)sin(lim0x
Solution
Find:)8sin()3tan(
xxlim
0x
=
=
=
=
11.1.3.1.
81
=83
=
00
)8sin()3tan(
xxlim
0x
)3cos(1.)3sin(.
)8sin( xxx
xx
)3cos(1).3sin(.
)8sin(1
xx
xlim0x
lim0x
)3cos(1.
3)3sin(.3.
)8sin(8.
81
xxx
xxlim
0x
Solution
1 )Find )2
(
)2
sin(
t
tlim2t
)2
(
)2
sin(
t
t=
00lim
2t
Put = t- so as t , 0. 2 2
= 1)2
(
)2
sin(
t
t=
lim2t
)sin(lim0
Solution
Derivative of y = sin xDerivative of y = sin x
sin cosd
x xdx
WHY?WHY?
Derivative of y = sin xDerivative of y = sin x
0( ) ( )
( ) limhdy f x h f x
f xdx h
0sin( ) sin( )
( ) limhx h x
f xh
0sin( )cos( ) cos( )sin( ) sin( )
limhx h x h x
h
0sin( )cos( ) sin( ) cos( )sin( )
limhx h x x h
h
0sin( )(cos( ) 1) cos( )sin( )
limhx h x h
h h
= 0 +cos(x)*1 = cos (x)
Derivative of Sine, CosineDerivative of Sine, Cosine
cos sind
x xdx
sin cosd
x xdx
Examples:
f(x) = x2 +sin(x)
f’(x) = 2x + cos(x)
•f(t) = cos(t) – 5t -2 , then f '(t) = -sin (t) +10t -3
Derivative of Sine, Cosine
Find tan x dxd
dxd tan x = dx
dxx
cossin
xxxxx
2cossin.sincos.cos =
2 2cos sin 12 2cos cos
x x
x x
=
Solution
Derivatives of Trigonometric Derivatives of Trigonometric FunctionsFunctions
sin cosd
x xdx
cos sind
x xdx
2tan secd
x xdx
2cot cscd
x xdx
sec sec tand
x x xdx
csc csc cotd
x x xdx
Find if: dxdy
1 (Y= xxx secsin32
xxxxdxdy tanseccos32
Solution
2 )Y = xxx csc.tancos
)2secsin(csc)cotcsc).(tan(cos xxxxxxxdxdy
Find y’ if y = xx
sin1cos
Solution
21( ) 5sin sec tan 7 3
2f x x x x x x
21( ) 5cos sec tan sec tan (1) 14
2f x x x x x x x x
1 sin( )
cos
xf x
x x
2
( cos ) (1 sin ) (1 sin ) ( cos )( )
( cos )
d dx x x x x x
dx dxf xx x
2
( cos )(cos ) (1 sin )(1 sin )( )
( cos )
x x x x xf x
x x
2 2 2 2
2 2
( cos cos ) (1 sin ) cos cos 1 sin( )
( cos ) ( cos )
x x x x x x x xf x
x x x x
2
cos( )
( cos )
x xf x
x x
Find :
=
0tancos)0sec(2
= 13
= 3
xx
tancossec2
lim0x
xx
tancossec2
lim0x
Solution
Has a tangent at x = 0 .
Show that the function:
F)x)=
0
)1sin(.2 xx X 0
X = 0
Find the values of a and b s.t the function:
F)x) =
bax
x 12 ,x 2
,x < 2
Is diff . Every where.