Book Mark Book Mark delicious
Limits Laws Tutor & Homework Help
Home > Online Tutoring > Algebra > Limits Laws
Tutors on NetTutors on Net
Limits Laws Tutoring Online, Tutor Help With Limits Laws:
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.


Online Limits Laws Help:

If you are stuck with a Limits Homework problem and need help, we have excellent tutors who can provide you with Homework Help. Our tutors who provide Limits Laws help are highly qualified. Our tutors have many years of industry experience and have had years of experience providing Limits Laws Homework Help. Please do send us the Limits Laws problems on which you need Help and we will forward then to our tutors for review.


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  = 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


Online Limits Laws Tutor Limits:

We have the best tutors in math in the industry. Our tutors can break down a complex Limits Laws problem into its sub parts and explain to you in detail how each step is performed. This approach of breaking down a problem has been appreciated by majority of our students for learning Limits Laws concepts. You will get one-to-one personalized attention through our online tutoring which will make learning fun and easy. Our tutors are highly qualified and hold advanced degrees. Please do send us a request for Limits Laws tutoring and experience the quality yourself.