g(x) lim exist (meaning they are finite numbers). 4. lim x a [f(x)g(x)] = lim x a f(x) lim x a g(x); (the limit of a sum is the sum of the limits).
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1 Click on tis symbol Lecture 4 : Calculating Limits using Limit Laws to view an interactive demonstration in Wolfram Alpa Using te definition of te it, a f(), we can derive many general laws of its, tat elp us to calculate its quickly and easily Te following rules apply to any functions f() and g() and also apply to left and rigt sided its: Suppose tat c is a constant and te its eist (meaning tey are finite numbers) f() and g() a a Ten a [f() + g()] = a f() + a g() ; (te it of a sum is te sum of te its) 2 a [f() g()] = a f() a g() ; (te it of a difference is te difference of te its) 3 a [cf()] = c a f(); (te it of a constant times a function is te constant times te it of te function) 4 a [f()g()] = a f() a g(); (Te it of a product is te product of te its) 5 a f() g() = a f() a g() if a g() 0; (te it of a quotient is te quotient of te its provided tat te it of te denominator is not 0) Eample If I am given tat f() = 2, 2 find te its tat eist (are a finite number): (a) g() = 5, 2 2f() + () = 2(2f() + ()) 2 g() 2 g() = 2 2 f() + 2 () 2 g() () = 0 2 = 2(2) since 2 g() 0 = 4 5 (b) f() 2 () (c) f()() 2 g() Note If a g() = 0 and a f() = b, were b is a finite number wit b 0, Ten: te values of te quotient f() can be made arbitrarily large in absolute value as a and tus g()
2 te it does not eist If te values of f() are positive as a in te above situation, ten g() a f() =, g() If te values of f() are negative as a in te above situation, ten g() a f() =, g() If on te oter and, if a g() = 0 = a f(), we cannot make any conclusions about te it Eample Find π cos π As approaces π from te left, cos approaces a finite number As approaces π from te left, π approaces 0 Terefore as approaces π from te left, te quotient cos approaces in absolute value π Te values of bot cos and π are negative as approaces π from te left, terefore π cos π = More powerful laws of its can be derived using te above laws -5 and our knowledge of some basic functions Te following can be proven reasonably easily ( we are still assuming tat c is a constant and a f() eists ); 6 a [f()] n = [ a f() ] n, were n is a positive integer (we see tis using rule 4 repeatedly) 7 a c = c, were c is a constant ( easy to prove from definition of it and easy to see from te grap, y = c) 8 a = a, (follows easily from te definition of it) 9 a n = a n were n is a positive integer (tis follows from rules 6 and 8) 0 a n = n a, were n is a positive integer and a > 0 if n is even (proof needs a little etra work and te binomial teorem) a n f() = n a f() assuming tat te a f() > 0 if n is even (We will look at tis in more detail wen we get to continuity) Eample Evaluate te following its and justify eac step: (a) (b) 3 + 2
3 (c) Determine te infinite it (see note above, say if te it is, or DNE) + 2 ( 2) Polynomial and Rational Functions Please review te relevant parts of Lectures 3, 4 and 7 from te Algebra/Precalculus review page Tis demonstration will elp you visualize some rational functions: Direct Substitution (Evaluation) Property If f is a polynomial or a rational function and a is in te domain of f, ten a f() = f(a) Tis follows easily from te rules sown above (Note tat tis is te case in part (a) of te eample above) if f() = P () is a rational function were P () and Q() are polynomials wit Q(a) = 0, ten: Q() P () If P (a) 0, we see from note above tat a = ± or DNE and is not equal to ± Q() If P (a) = 0 we can cancel a factor of te polynomial P () wit a factor of te polynomial Q() and te resulting rational function may ave a finite it or an infinite it or no it at = a P () Te it of te new quotient as a is equal to a by te following observation wic Q() we made in te last lecture: Note 2: If () = g() wen a, ten a () = a g() provided te its eist Eample to ± : Determine if te following its are finite, equal to ± or DNE and are not equal (a) (b) 2 6 (c) Wic of te following is true: 2 6 = +, =, ±, DNE and is not 3
4 Eample Evaluate te it (finis te calculation) 0 (3 + ) 2 (3) 2 0 (3+) 2 (3) 2 = = Eample Evaluate te following it: Recall also our observation from te last day wic can be proven rigorously from te definition (tis is good to keep in mind wen dealing wit piecewise defined functions): Teorm a f() = L if and only if a f() = L = a + f() Eample Evaluate te it if it eists: Te following teorems elp us calculate some important its by comparing te beavior of a function wit tat of oter functions for wic we can calculate its: 4
5 Teorem If f() g() wen is near a(ecept possible at a) and te its of f() and g() bot eist as approaces a, ten f() g() a a Te Sandwic (squeeze) Teorem If f() g() () wen is near a (ecept possibly at a) and f() = () = L a a ten g() = L a Recall last day, we saw tat 0 sin(/) does not eist because of ow te function oscillates near = 0 However we can see from te grap below and te above teorem tat 0 2 sin(/) = 0, since te grap of te function is sandwiced between y = 2 and y = 2 : 05 K K K05 Eample Calculate te it 0 2 sin K We ave sin(/) O for all, multiplying across by 2 (wic is positive), we get 2 2 sin(/) 2 for all, Using te Sandwic teorem, we get 0 = 0 2 o 2 sin(/) 0 2 = 0 Hence we can conclude tat 0 2 sin(/) = 0 Eample Decide if te following it eists and if so find its values: o 00 cos 2 (π/) 5
6 Etra Eamples, attempt te problems before looking at te solutions Decide if te following its eist and if a it eists, find its value () (2) ( 2) 2 (3) 0 ( ) (4) (5) (6) If 2 g() for all, evaluate g() (7) Determine if te following it is finite, ± or DNE and is not ± ( 3)( + 2) ( )( 2) 6
7 Etra Eamples, attempt te problems before looking at te solutions Decide if te following its eist and if a it eists, find its value () Since tis is a polynomial function, we can calculate te it by direct substitution: = 4 + 2() = 7 (2) ( 2) 2 Tis is a rational function, were bot numerator and denominator approac 0 as approaces 2 We factor te numerator to get After cancellation, we get ( )( 2) = 2 ( 2) 2 2 ( 2) 2 ( )( 2) ( ) = 2 ( 2) 2 2 ( 2) Now tis is a rational function were te numerator approaces as 2 and te denominator approaces 0 as 2 Terefore ( ) 2 ( 2) does not eist We can analyze tis it a little furter, by cecking out te left and rigt and its at 2 As approaces 2 from te left, te values of ( ) are positive (approacing a constant ) and te values of ( 2) are negative ( approacing 0) Terefore te values of ( ) are negative and ( 2) become very large in absolute value Terefore Similarly, you can sow tat and terefore te grap of y = ( ) ( 2) (ceck it out on your calculator) (3) 0 ( ) Let f() = ( ) 2 ( 2) = ( ) 2 ( 2) = +, as a vertical asymptote at = 2 We write tis function as a piecewise defined function: = 0 > 0 f() = + = 2 0 7
8 0 ( ) eists only if te left and rigt and its eist and are equal 0 +( ) = = 0 and 0 ( ) = 0 2 = Since te its do not matc, we ave ( 0 ) DNE (4) Since = 0 0, we ave Now = 0 0 ( ) = 0 0 > 0 = 0 Clearly 0 + = 0 = 0 Hence 0 = 0 and (5) = 0 0 = 0 0 = 0 Since = 0 = 0, we cannot determine weter tis it eists or not from te it laws witout some transformation We ave ( = )( ) ( ) ( 4 + ) 2 4) 0 ( ) = = 0 (4 + ) 4 ( ) = 0 ( ) = 0 (6) If 2 g() for all, evaluate g() ( ) = 4 We use te Sandwic teorem ere Since 2 g() 2 + 2, we ave 2 g() (2 + 2), terefore and ence 2 g() 2 g() = 2 8
9 (7) Determine if te following it is finite, ± or DNE and is not ± ( 3)( + 2) ( )( 2) Let P () = ( 3)( + 2) and Q() = ( )( 2) We ave P () = 6 0 and Q() = 0 Terefore te values of P () = ( 3)(+2) get larger in absolute value as approaces Q() ( )( 2) As approaces from te left, ( 3) < 0, ( 2) < 0, ( ) < 0, and ( + 2) > 0, terefore te quotient ( 3)(+2) < 0 as approaces from te left and terefore ( )( 2) ( 3)( + 2) ( )( 2) = 9
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