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Calculating the Limits of sin(x) and cos(x) using Different Methods - Prof. Marwan Zabdawi, Assignments of Calculus

The steps to calculate the limits of the functions sin(x) and cos(x) as x approaches 0 using numerical, geometric, and algebraic methods. It also includes the use of fermat's definition of derivatives and the squeeze theorem. Taken from a university mathematics course, specifically math 1501.

Typology: Assignments

Pre 2010

Uploaded on 08/04/2009

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16 June 2007
Martina McCauley
Math 1501
Derivative of
x
xsin
Show that
1
sin
lim
0
x
x
x
a) Numerically
b) Geometrically
c) Algebraically
a) Numerically
(
x
is in radians)
x
x
xsin
0.1 0.998
0.01 1.00
0.001 1.000
--------------------------------
-0.1 0.998
-0.01 1.000
-0.001 1.000
1
sin
lim
0
x
x
x
pf3
pf4
pf5

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Download Calculating the Limits of sin(x) and cos(x) using Different Methods - Prof. Marwan Zabdawi and more Assignments Calculus in PDF only on Docsity!

16 June 2007

Martina McCauley

Math 1501

Derivative of

x

sin x

Show that

sin

lim

0

x

x

x

a) Numerically

b) Geometrically

c) Algebraically

a) Numerically

x

is in radians)

x

x

sin x

 1

sin

lim

0

x

x

x

b) Geometrically

The graph of

x

sin x

is

sin

lim

0

x

x

x

c) Algebraically

Area of

 OAB

< Area of sector OAB <

 OCB

2

hOB r OBBC

(cos( ) 1 )

(cos 1 )

lim

2

0

h h

h

h

But h h

2 2

cos  1 sin , because cos sin 1

2 2

hh

(cos( ) 1 )

sin

lim

cos( ) 1

lim

2

0 0

 

h h

h

h

h

h h

  

cos( ) 1

lim

sin( )

lim( sin( )) lim

0 0 0

h h

h

h

h h h

QED

Show using Fermat’s Definition of Derivatives that:

x

dx

d x

cos

sin

Let

f ( x )sin x

h

f x h f x

dx

df

h

lim

0

h

x h x

h

sin( ) sin

lim

0

h

x h x h x

h

sin cos( ) cos sin( ) sin

lim

0

h

x h x h

h

sin (cos( ) 1 ) cos sin( )

lim

0

h

x h

h

x h

h h

cos sin( )

lim

sin (cos( ) 1 )

lim

 0  0

cos( ) 1

lim

0

h

h

h

h

h

x

h

h

x

h h

sin( )

cos lim

(cos( ) 1 )

sin lim

 0  0

But from Eq. (1) we have

sin( )

lim

0

h

h

h

And from Eq. (2) we have

(cos( ) 1 )

lim

0

h

h

h

x x x

dx

df

sin  0 cos  1 cos

For

f ( x )sin x

x

dx

df

f ( x ) cos

QED