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Laws Of Limits

Limits Laws Assignment / Homework Help
We can calculate the limits of a function from its graph as shown in the following example.

Illustrative Example

Online Calculating Limits

All the following statements about the function in the graph above are true.

At x = -2 :     lim    f (x) = 5
x -> -2+
At x = 0 :     lim    f (x) = 1
x -> 0-
    lim    f (x) = 1
x -> 0+
    lim    f (x) = 1
x -> 0
Also f (1) = 1
At x = 1 :     lim    f (x) = 1
x -> 1-
Even though f (1) = 2
    lim    f (x) = 2
x -> 1+
Hence     lim    f (x)   does not exist as the One-sided limits are not equal.
             x -> 1+
At x = 3 :     lim    f (x)  =  6
x -> 3-
=     lim    f (x)
     x -> 1+
Hence     lim    f (x)    = 6 even though f (x) = 3
             x -> 3
At x = 5 : Hence     lim    f (x)    = -2
             x -> 5-

At every other point 'a' between -2 and 5, f (x)  has a limit as x -> a.


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Limit Laws

Suppose 'c'  is a constant and also  lim  f (x)  and  lim  g (x)  both exist then the following rules hold true in the calculation of limits.
                                                       x -> a              x -> a

Rules:   lim     = c      and      lim    x = a
x -> a                           x -> a
Addition Rule:   lim  [  f ( x ) + g ( x )  ]   =    lim    f ( x )      +      lim     g ( x )
x -> a                                    x -> a                        x -> a
Difference Rule:   lim  [  f ( x ) - g ( x )  ]   =    lim    f ( x )      -      lim     g ( x )
x -> a                                    x -> a                        x -> a
Scalar Product Rule:   lim  [  c  f ( x )  ]   = c  [  lim    f ( x ) ]
x -> a                               x -> a
Product Rule:   lim  [  f ( x ) . g ( x )  ]   =    lim    f ( x )      .      lim     g ( x )
x -> a                                    x -> a                        x -> a
Quotient Rule:
lim
x -> a
  f ( x )



 g ( x )
  =   lim     f ( x )
x -> a

  lim     g ( x )
x -> a
  if   lim  g ( x ) ≠ 0
     x -> a

From the Product rule we can also conclude by repeated multiplication that:
(Power Rule)
lim      [ f ( x )  ]n
x -> a
= [   lim      f ( x )  ]n
    x -> a
where n is positive integer.
Similarly we have
(Root Rule)
 lim      n f ( x ) 
x -> a
= n  lim      f ( x ) 
        x -> a
where n is positive integer and if it is even then we assume that  lim   f ( x ) > 0.
        x -> a

Illustrative Examples:

  •  lim      2
    x -> 2
    = (  lim      x )
       x -> 2
    (  lim      x )
       x -> 2
    ............. Product Rule
    = ( 2 ) ( 2 ) ............. Basic Substitution Rule
    = 4
    Therefore,
     lim      2
    x -> 2
    = 4
  •   lim    3x2    =    3 lim    x2    =    3 ( 4 )  =  12
    x -> 2                  x -> 2

  • lim
    x -> 2
      2 - 2



     3x2
    =   lim    ( x2 - 2 )
    x -> 2

      lim     3x2
    x -> 2
    .................. (Quotient Rule)

    = lim    x2   -    lim     2
    x -> 2           x -> 2

    12
       ............... (Subtraction Rule)

    = 4 - 2  
      12  
       ............... (Basic Rule)

    =   1
      6  

    Conclusion: (Direct Substitution Property)

    If f (x) is any polynomial function then

    lim    f (x)    =  f (a)
    x -> 2

    Also if  g (x)  is also a polynomial and  g (a)  ≠  0 then

    lim
    x -> a
      f ( x )



     g ( x )
      =   f ( a )



     g ( a )

Examples:

  • lim
    x -> 1
      x3 + 4x2 + 3x + 2



              6x - 1
     =   13 + 4 (1)2 + 3 (1) + 2



               6 (1) - 1
     = 10



    5
     = 2

  • lim
    x -> 2
      x2 - 4



     x - 2

    Here we cannot apply the substitution as it takes 0 / 0 form nor can we apply quotient rule as the denominator x - 2 = 0 for x = 2.
    Instead let us apply some algebra before taking the limit.
      x2 - 4



     x - 2
    =   (x + 2) (x - 2)



     (x - 2)

    Now we know that x -> 2 but x ≠ 2. Hence x - 2 ≠ 0 and we can cancel it off.
    lim
    x -> 2
      x2 - 4



     x - 2
    = lim
    x -> 2
    (x + 2)  (x - 2) 



     (x - 2) 
    = lim    (x + 2)
    x -> 2
    =   2   +   2   =   4

    Therefore,
    lim
    x -> 2
      x2 - 4



     x - 2
    = 4


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