# Limit-Definition-of-the-Derivative

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```					                         Limit Definition of the Derivative
You won't have to calculate the derivative using def of derivative. You should recognize its
form, then take a derivative of the function by another method.

FORM A
f ( x  h)  f ( x)
Definition of the Derivative           f ( x)  lim
h0             h

Example 1:     lim     ( x  h) 2  x 2
?
h0            h

Answer: 2x. Recognize that this is definition of the derivative. The function is x2. Now take
the derivative by power rule and find that d/dx of x2 is 2x.

Example 2:     lim      tan( x  h)  tan( x)
?
h0                h

Answer: sec2 (x)           f(x) = tan(x)               f ' (x) = sec2 (x)

Example 3:     lim        ( x  h) 4  x 4
h0                         ?
h
Answer: 4x3            f(x) = x4            f ' (x) = 4x3

Example 4:     lim      2( x  h) 2  3( x  h)  2 x 3  3x
?
h0                        h

Answer: 4x+3           f(x) = 2x2+3x         f ' (x) = 4x+3

sin( 2( x  h))  sin 2 x
Example 5:     lim                                  ?
h0                    h

Answer: 2 cos 2x        f(x) = sin2x         f ' (x) = 2 cos 2x

cos( x  h) 2  cos x 2
Example 6:     lim                                 ?
h
h0

Answer: - 2x sin(x2)      f(x) = cos (x2)       f ' (x) = - 2x sin(x2)
FORM B
Definition of the Derivative at a point (x = c)         lim     f ( x )  f (c )
 f ' (c )
xc          xc

Example 7:     lim       x2  9
?
x3        x 3

Answer: 6 Recognize that this is definition of the derivative at a point. The function is x2 and
it is evaluated at the point x=3. Now take the derivative by power rule and find that d/dx of x2 is
2x, then plug in x=3 and get 2(3) = 6.

Example 8:     lim        x 2  52
?
x5          x 5

Answer: 10             f(x) = x2        x =5 is the x value desired       f ' (x) = 2x        plug in x=5
f’(5) = 10

trickier version of example 8
Example 9: lim              x 2  25
x5                  ?
x5
Answer: 10             This time the 52 has been given in a different form so it appears more
difficult but it is the same as Example 8.     f(x) = x2 x =5 is the x value desired
f ' (x) = 2x        plug in x=5 f’(5) = 10

Example 10: lim                         
sin x  sin( )
xπ/3                       3 ?

x
3

Answer: 1/2          f(x) = sin(x)      x =pi/3   f ' (x) = cos (x). f '(pi/3) = cos(pi/3) = 1/2

trickier version of Example 10
Example 11: lim                      3
xπ/3      sin x 
2 ?

x
3

Answer: 1/2         This time the sin(π/3) has been given in a different form so it appears more
difficult but it is the same as Example 10.         √(3)/2 is given in place of sin(pi/3)
f(x) = sin(x) x = π/3 so f ' (x) = cos (x) f ' (π/3) = cos(π/3) = 1/2
another tricky one:
Example 12: lim                              2
xπ/4            cos x 
2 ?

x
4

cos x  cos( )
Answer: - √(2)/2       Recognize this as           lim                        4

x π/4              x
4
f(x) = cos(x)      x = π/4         so f ' (x) = - sin(x)           f ' (π/4 ) = - sin (π/4 ) = - √(2)/2

(Ostebee Zorn 2nd ed. section 2.7, page 148)
Example 13: lim       sin x
x0           ?
x

Answer: 1       Recognize this as            lim    sin x sin x  sin 0
x0         
x       x0
since sin(0) = 0             f(x) = sin(x)    x=0          f ' (x) = cos (x)        f ' (0) = cos(0) = 1

Example 14: lim           sin x
xπ                 ?
x 
Answer: -1         Recognize this as         lim    sin x sin x  0 sin x  sin 
         
xπ    x     x         x 

since sin π = 0           f(x) = sin(x)       x=π          f ' (x) = cos (x)        f ' (π) = cos(π) = -1

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