Sections 1.3 Computation of Limits

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1 1 Sections 1.3 Computation of Limits We will shortly introduce the it laws. Limit laws allows us to evaluate the it of more complicated functions using the it of simpler ones. Theorem Suppose that c is a constant and the its f() and g() eists. Then; 1) 2) 3) 4) 5) [f() ± g()] = f() ± g() [cf()] = c f() [f()g()] = f() g() If g() 0, then [ f() f() g()] = g() [f()]n = [ f()] n n is a positive integer Indeed the same rules hold for directional its, so that (for instance) if f() and g() eists, then + g()] = f() + [f() g() Corollary If f() is a polynomial, then f() = f(a). Proof: If f() is a polynomial, we know that f() = a n n + a n 1 n a 1 + a 0. Then we have

2 2 f() = a n n + a n 1 n a 1 + a 0 = [a n n ] + [a n 1 n 1 ] [a 1 ] + [a 0 ] = a n [ n ] + a n 1 [ n 1 ] a 1 [] + a 0 = a n ( ) n + a n 1 ( ) n a 1 ( ) + a 0 = a n a n + a n 1 a n a 1 a + a 0 = f(a) Eample Find the 5 ( ). We will use the same steps as we did in the proof of the above Corollary. 5 ( ) = by Law Eample Find the = by Law = 2[ ] by Law = 2(5) 2 3(5) + 4 = First note that 5 3 = 5 3 Law 4 for quotients. =to 2 = So we may use the = = ( 2)3 + 2( 2) = 1 11 Theorem If f is a rational function and a is in the domain of f, then f() = f(a). After the Eample above the proof of this theorem should be clear. Let s see what happens if a is not in the domain of our rational function.

3 3 Eample Find We cannot simply substitute = 0 since f(0) is not defined. Hence we cannot apply the Law 4 for quotients (denominator is zero). However we can rationalize the numerator as follows: = ( 2 + 9) 9 = 2 ( ) = 2 2 ( ) 1 = ( ) = 1 ( 2 + 9) + 3 Eample Find = = 1 6 Note that again we cannot use the Law 4 for quotients because the denominator is zero at = 2. So we need another algebra trick; = ( 2)( + 2) ( 2) = ( + 2) = 4 In the above eample note that f() + 2. The rational function is very much like + 2 ecept it is not defined at = 2. So we are using the power of the it. Theorem Let f() = L and n is any positive integer, then If n is even, we need f() > 0. n f() = n f()

4 4 Eample Find = ( ) = ( 2) 2 + 3( 2) + 6 = 4 Eample Find 2. Your first reaction to this problem should be that it does not eits. Because any open interval around =2 contains values of not in the domain of 2. But if we correct the question and discuss the it from the right we have an answer: 2 = 0. If you consider 2 does not eist. By a fact we + have seen before you can also conclude that 2 does not eist. Some Cautionary Eamples: Note that I have underlined the verb eists in the statement of the it laws. You have to be cautious of this fact when you do the it calculations. If either one of the its do not eists, then you cannot automatically assume that you are able to use these rules. The Limit Laws guarantee results only if both results eits. So natural question might come to your mind such as: Question Does [f()+g()] eits even though f() and g() does not eists? Answer It depends on f() and g(). Check out the following: { 0 < 0 Let f() = 1 0 { 1 < 0 and g() = 0 0 Both f() and g() do not eist because both of their one-sided its are not equal. f() = 1 f() = 0 and g() = g() = 1 But f() + g() = 1 for all. So [f() + g()] = 1 hence the it of the sum eists.

5 Question Does f()g() eist even though f() eists but g() does not. Answer It again depends on f() and g(). Take your f() = and g() = 1. = 0 and 1/ does not eist. But 1/ = 1 and eists. (Note that 1/ is not defined at = 0 because 1/ is not.) Question What is [3f() + 2g()] where f() is the blue one and g() is the red. 5 Note again we cannot use the it laws because g() does not eist. So instead note that the directional its eist and hence you can use the it laws for the directional its and compare them for the eistence of [3f() + 2g()]. (Recall: [3f() + 2g()] eists iff + 2g()] and [3f() +[3f() + 2g()] eists and they are equal.) Now: f() = f() = 1, g() = 2 and + Using our it laws for directional its, we have and g() = g()] = + = ( 2) = 1 [3f() [3f()] [2g()] + 2g()] = + = ( 1) = 1 +[3f() +[3f()] +[2g()] Since these two directional it do not agree, we conclude that [3f() + 2g()] does not eist. So that you have more fun while solving the hw problems and in discussion sessions I ll state the following result from your book without proof.

6 6 Theorem For any number a, we have sin() = sin(a) cos() = cos(a) e = e a ln() = ln(a), for a > 0 If p() is a polynomial, and then f(p()) = L f() = L p(a) sin 1 () = sin 1 (a) for 1 < a < 1 cos 1 () = cos 1 (a) for 1 < a < 1 tan 1 () = tan 1 (a) for < a < We will re-visit the results of this theorem when we deal with continuity in the net section. Now let s learn another tool that will help us to evaluate some important its such as sin() Theorem If f() g() when is near a (ecept possibly at a) and the its of f and g eists as approaches a, then f() g() Squeeze Theorem/Sandwich Theorem If f() g() h() when is near a (ecept for possibly at a) and f() = h() = L then, g() = L Eample Show that 2 sin( 1 ) = 0. First note that you cannot use the it law 3 for products of functions because sin(1/) does not eist. However, since 1 sin( 1 ) 1 for all 0 we have 2 2 sin( 1 ) 2 since 2 0 always.

7 7 Also observe this relation in the graph below: Out[22]= We know that 2 = 2 = 0. Taking f() = 2, g() = 2 sin( 1 ) and h() = 2 in the Squeeze Theorem, we obtain 2 sin( 1 ) = 0 Little Eercise Try yourself to show sin(1/) = 0. Here you will need your directional it abilities along with the Squeeze Theorem.