26 the problem now is that we have two deltas and the

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Chapter 4 / Exercise 16
Algebra and Trigonometry: Real Mathematics, Real People
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(26) The problem now is that we have two deltas, and the definition of the limit requires that we give a single delta. The way to solve this is to recall that if we have found a delta which works, any smaller delta will also work. Hence by setting δ = min { δ f , δ g } we can guarantee that δ δ f and δ δ g so that both equations in ( 26 ) hold. Hence if 0 < | x - c | < δ then [ f ( x ) + g ( x )] - [ L + M ] | f ( x ) - L | + | g ( x ) - M | < 2 + 2 = . This is what we wanted to show, and so the proof is complete. Exercise 2.25. Complete the proofs for Theorem 2.24 . We can immediately use this to prove some very powerful and useful results. Corollary 2.26 If f ( x ) = p ( x ) q ( x ) is any rational functions (so that p ( x ) and q ( x ) are polynomials), and c 2 R is such that q ( c ) 6 = 0 then lim x ! c p ( x ) q ( x ) = p ( c ) q ( c ) . Proof. The key is to first show this for polynomials and apply the limit laws. It is easy to see (the student should check!) that lim x ! c x = c and so by induction, (the student can show that) lim x ! c x n = c n . Let p ( x ) = a n x n + a n - 1 x n 1 + · · · + c 1 x + c 0 be an arbitrary polynomial. Our discussion tells us that we know how to deal with every occurrence of the function x 7! x , and so using the Theorem 2.24 c 2015 Tyler Holden 41
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Algebra and Trigonometry: Real Mathematics, Real People
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Chapter 4 / Exercise 16
Algebra and Trigonometry: Real Mathematics, Real People
Larson
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2 Limits 2.5 Limits at Infinity we have lim x ! c p ( x ) = lim x ! c [ a n x n + · · · + a 1 x + a 0 ] = a n h lim x ! c x i n + a n - 1 h lim x ! c x i n - 1 + · · · + a 1 h lim x ! c x i + a 0 = a n c n + a n - 1 c n - 1 + · · · + a 1 c + a 0 = p ( c ) . Thus the result holds for any polynomial. Now if p ( x ) and q ( x ) are two polynomials and q ( c ) 6 = 0, then the limit laws for quotients implies lim x ! c p ( x ) q ( x ) = lim x ! c p ( x ) lim x ! c q ( x ) = p ( c ) q ( c ) , as required. 2.5 Limits at Infinity There are two notions of infinity that we will be interested in tackling, both with definitions similar to the - δ definition we saw before. Definition 2.27. We say that lim x !1 f ( x ) = L if for every > 0 there exists an M 2 R such that whenever x > M then | f ( x ) - L | < . We say that lim x ! c f ( x ) = 1 if for every M 2 R there exists a δ > 0 such that if 0 < | x - c | < δ then f ( x ) > M . In both cases, to consider -1 we replace x > M, f ( x ) > M with x < M, f ( x ) < M . Exercise 2.28. Determine an appropriate statement for lim x !1 f ( x ) = 1 . Example 2.29. Show that lim x !1 x x + 1 = 1. 42 c 2015 Tyler Holden
2.5 Limits at Infinity 2 Limits Solution. Let > 0 be given and take M = max - 1 , - 1 - 1 . If x > M then since x M ≥ - 1 we know that | x + 1 | = x + 1. Furthermore, since x > - 1 - 1 we know that x + 1 > 1 , thus x x + 1 - 1 = - 1 x + 1 = 1 | x + 1 | < as required. Example 2.30. Show that lim x ! 0 1 x 2 = 1 . Solution. Let M 2 R be given and let δ = 1 p | M | . If | x | < δ = 1 p | M | then 0 < x 2 < 1 | M | and so f ( x ) = 1 x 2 > 1 | M | > 1 M . Just as it was possible to use the - δ definition to show that finite limits do not exist, we can use our rigorous definitions to show that limits at infinity do not exist. Indeed, we say that lim x !1 f ( x ) does not exist if for every L 2 R there exists an > 0 such that for every M 2 R there exists an x such that x > M and | f ( x ) - L | .

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