2311lectureoutline2.3

2311lectureoutline2.3 - ( ) lim x a f x L = if and only if...

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MAC2311, Calculus I 2.3 Calculating Limits Using Limit Laws Properties of Limits (Limit Laws): Suppose that c is a constant and the limits lim ( ) x a f x and lim ( ) x a g x exist. Then… 1. ( 2. 3. ( 4. , 5. ( ) ( )) ( ) ( ) ( ) ( ) ( ) ( )) ( ) ( ) ( ) ( ) ( ) 0 ( ) ( ) ( ) ( ) lim lim lim lim lim lim lim lim lim lim lim lim lim lim n n x a x a x a x a x a x a x a x a x a x a x a x a x a x a aslong as g c g g g f x g x f x x f x c f x f x g x f x x f x f x x g x x f x f x ± ± = ² ³ ² ³ ± = ± = = = Additional special limits include: lim lim x a x a c c x a = = Methods for finding limits: 1. Use direct substitution. 2. Rewrite/manipulate the function, THEN use direct substitution. 3. Look at the graph of the function. Theorem: The
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Unformatted text preview: ( ) lim x a f x L = if and only if ( ) lim x a f x L + = = ( ) lim x a f x- . Theorem: If ( ) ( ) f x g x when x is near a (except possibly at a ) and the limits of f(x) and g(x) both exist as x approaches a , then ( ) ( ) lim lim x a x a f x g x . The Squeeze Theorem: If ( ) ( ) ( ) f x g x h x when x is near a (except possibly at a ) and ( ) ( ) lim lim x a x a h f x x L = = , then ( ) lim x a g x L = . Try Exercises 2, 4, 8, 14, 18, 22, 26, and 46 on pages 106-107 in the textbook....
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