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 x→a f( x )       lim x→a g( x )


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


2. lim x→a [ f( x )−g( x ) ]= lim x→a f( x )− lim x→a g( x )      Difference of Limits


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


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


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


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


7. lim x→a c=c


8. lim x→a x=a


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


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


11. lim x→a f( x ) n = lim x→a f( x ) n      where n is a positive integer and if n is even, we assume that lim x→a 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 x→2 ( 2 x 2 −2 )( x 3 −x+5 )

Step 1: Apply the Product of Limits Law 4.


lim x→a [ f( x )g( x ) ]= lim x→a f( x )· lim x→a g( x )

lim x→2 ( 2 x 2 −2 )· lim x→2 ( x 3 −x+5 )

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


Law 2: lim x→a [ f( x )−g( x ) ]= lim x→a f( x )− lim x→a g( x )


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

[ lim x→2 2 x 2 − lim2 x→2 ] · [lim x→2   x 3 − lim x→2  x+ lim x→2 5 ]

Step 3: Apply the Constant Multiple Law 3.


lim x→a [ cf( x ) ]=c lim x→a f( x )

[2  lim x→2 x 2 − lim2 x→2 ] · [lim x→2   x 3 − lim x→2  x+ lim x→2  5 ]

Step 4: Apply Limit Law 9.


lim x→a x n = a n

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

Step 5: Apply Limit Law 7.


lim x→a c=c

[2( 2 2 )−2] · [ 2 3 − lim x→2  x+5]

Step 6: Apply Limit Law 8.


lim x→a x=a

[2( 2 2 )−2] · [ 2 3 −2+5]

Step 7: Evaluate the limit as x→2 .

[2( 2 2 )−2] · [ 2 3 −2+5]=66


lim x→2 ( 2 x 2 −2 )( x 3 −x+5 )=66

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

lim x→8 ( 3−5 x 2 + x 3 )( 2+ x 3 )

Step 1: Apply the Product of Limits Law 4.


lim x→a [ f( x )g( x ) ]= lim x→a f( x )· lim x→a g( x )

lim x→8 ( 3−5 x 2 + x 3 )· lim x→8 ( 2+ x 3 )

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


Law 2: lim x→a [ f( x )−g( x ) ]= lim x→a f( x )− lim x→a g( x )


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

[lim x→8 3− lim x→8  5 x 2 + lim x→8   x 3 ] · [ lim x→8  2+ lim x→8   x 3 ]

Step 3: Apply the Constant Multiple Law 3.


lim x→a [ cf( x ) ]=c lim x→a f( x )

[lim x→8 3−5  lim x→8   x 2 + lim x→8   x 3 ] · [ lim x→8  2+ lim x→8   x 3 ]

Step 4: Apply Limit Law 7.


lim x→a c=c

[3−5  lim x→8   x 2 + lim x→8   x 3 ] · [2+ lim x→8   x 3 ]

Step 5: Apply Limit Law 9.


lim x→a x n = a n

[3−5 ( 8 2 )+ 8 3  ] · [2+ lim x→8   x 3 ]

Step 6: Apply Limit Law 11.


lim x→a f( x ) n = lim x→a f( x ) n

[3−5 ( 8 2 )+ 8 3  ] · [2+ 8 3 ]

Step 7: Evaluate the limit as x→2 .

[3−5 ( 8 2 )+ 8 3  ] · [2+ 8 3 ]=780


lim x→8 ( 3−5 x 2 + x 3 )( 2+ x 3 )=780





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


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