Limits: Limit Laws

Graphs and tables can be used to guess the values of limits but these are just estimates and these methods have inherent problems. A better method is to use the following properties of limits called Limit Laws.

LIMIT LAWS:


Assume the following limits exist and c is a constant.


lim xa f( x )       lim xa g( x )


1. lim xa [ f( x )+g( x ) ]= lim xa f( x )+ lim xa g( x )      Sum of Limits


2. lim xa [ f( x )g( x ) ]= lim xa f( x ) lim xa g( x )      Difference of Limits


3. lim xa [ cf( x ) ]=c lim xa f( x )     Constant Multiple


4. lim xa [ f( x )g( x ) ]= lim xa f( x )· lim xa g( x ) Product of Limits


5. lim xa [ f( x ) g( x ) ]= lim xa f( x ) lim xa g( x ) , g( x )0    Quotient of Limits


6. lim xa [ f ( x ) n ]= [ lim xa f( x ) ] n where n is a positive integer


7. lim xa c=c


8. lim xa x=a


9. lim xa x n = a n where n is a positive integer


10. lim xa x n = a n    where n is a positive integer and if n is even we assume a > 0


11. lim xa f( x ) n = lim xa f( x ) n      where n is a positive integer and if n is even, we assume that lim xa f( x )>0



Let's use these laws to calculate the limits in a couple examples.

Example 1: Evaluate the following limit and justify each step.

lim x2 ( 2 x 2 2 )( x 3 x+5 )

Step 1: Apply the Product of Limits Law 4.


lim xa [ f( x )g( x ) ]= lim xa f( x )· lim xa g( x )

lim x2 ( 2 x 2 2 )· lim x2 ( x 3 x+5 )

Step 2: Apply the Difference and Sum of Limit Laws 2 & 1.


Law 2: lim xa [ f( x )g( x ) ]= lim xa f( x ) lim xa g( x )


Law 1: lim xa [ f( x )+g( x ) ]= lim xa f( x )+ lim xa g( x )

[ lim x2 2 x 2 lim2 x2 ] · [lim x2   x 3 lim x2  x+ lim x2 5 ]

Step 3: Apply the Constant Multiple Law 3.


lim xa [ cf( x ) ]=c lim xa f( x )

[2  lim x2 x 2 lim2 x2 ] · [lim x2   x 3 lim x2  x+ lim x2  5 ]

Step 4: Apply Limit Law 9.


lim xa x n = a n

[2( 2 2 ) lim2 x2 ] · [ 2 3 lim x2  x+ lim x2  5]

Step 5: Apply Limit Law 7.


lim xa c=c

[2( 2 2 )2] · [ 2 3 lim x2  x+5]

Step 6: Apply Limit Law 8.


lim xa x=a

[2( 2 2 )2] · [ 2 3 2+5]

Step 7: Evaluate the limit as x2 .

[2( 2 2 )2] · [ 2 3 2+5]=66


lim x2 ( 2 x 2 2 )( x 3 x+5 )=66

Example 2: Evaluate the following limit and justify each step.

lim x8 ( 35 x 2 + x 3 )( 2+ x 3 )

Step 1: Apply the Product of Limits Law 4.


lim xa [ f( x )g( x ) ]= lim xa f( x )· lim xa g( x )

lim x8 ( 35 x 2 + x 3 )· lim x8 ( 2+ x 3 )

Step 2: Apply the Difference and Sum of Limit Laws 2 & 1.


Law 2: lim xa [ f( x )g( x ) ]= lim xa f( x ) lim xa g( x )


Law 1: lim xa [ f( x )+g( x ) ]= lim xa f( x )+ lim xa g( x )

[lim x8 3 lim x8  5 x 2 + lim x8   x 3 ] · [ lim x8  2+ lim x8   x 3 ]

Step 3: Apply the Constant Multiple Law 3.


lim xa [ cf( x ) ]=c lim xa f( x )

[lim x8 35  lim x8   x 2 + lim x8   x 3 ] · [ lim x8  2+ lim x8   x 3 ]

Step 4: Apply Limit Law 7.


lim xa c=c

[35  lim x8   x 2 + lim x8   x 3 ] · [2+ lim x8   x 3 ]

Step 5: Apply Limit Law 9.


lim xa x n = a n

[35 ( 8 2 )+ 8 3  ] · [2+ lim x8   x 3 ]

Step 6: Apply Limit Law 11.


lim xa f( x ) n = lim xa f( x ) n

[35 ( 8 2 )+ 8 3  ] · [2+ 8 3 ]

Step 7: Evaluate the limit as x2 .

[35 ( 8 2 )+ 8 3  ] · [2+ 8 3 ]=780


lim x8 ( 35 x 2 + x 3 )( 2+ x 3 )=780





Related Links:
Math
algebra
General Differentiation Rules
Exponential Differentiation Rules
Calculus Topics


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