    # Laws Of Limits We can calculate the limits of a function from its graph as shown in the following example.

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

•  lim x -> 2 x 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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