Fifth lecture for MAT137 - LEC5201 Topics for today: 1. More limits 2. Squeeze theorems 3. Continuity Required videos for next week: 2.21-2.22, 3.1-3.8 We will do some review for the test next lecture. Jeffrey Im MAT137 October 10, 2018 1 / 16
The course’s convention for when limits don’t exist Suppose f is a real-valued function with domain R and that a ∈ R . Nonexistence of limits We say that x → a f ( x ) does not exist lim when there is not an L ∈ R for which lim x → a f ( x ) = L . Limits which tend to infinity will be treated as limits not existing in a particular way. Jeffrey Im MAT137 October 10, 2018 2 / 16
Indeterminate form Let a ∈ R and let f and g be positive functions defined near a . Assume x → a f ( x ) = lim lim x → a g ( x ) = 0 . What can be concluded about f ( x ) lim g ( x )? x → a 1. The limit is 1. 2. The limit is 0. 3. The limit is ∞ . 4. The limit does not exist. 5. We do not have enough information to decide. Jeffrey Im MAT137 October 10, 2018 3 / 16
A theorem about limits Let f be a function with domain R such that x → 0 f ( x ) = 3 . lim Prove that x → 0 [5 f (2 x )] = 15 lim directly from the definition of the limit (No appeals to limit laws!). 1. Write down what needs to be proved. 2. Do some scratch work. 3. Write up a proof. Jeffrey Im MAT137 October 10, 2018 4 / 16
Proof feedback 1. Is the structure of the proof correct? (First fix ε , then choose δ , then ...) 2. Did you say exactly what δ is? 3. Is the proof self-contained? (I do not need to read the rough work) 4. Are all variables defined? In the right order? 5. Do all steps follow logically from what comes before? Do you start from what you know and prove what you have to prove? 6. Did you remember not to assume your conclusion? Jeffrey Im MAT137 October 10, 2018 5 / 16
Infinite version of the squeeze theorem The usual squeeze theorem Let a , L ∈ R and suppose f , g , h are functions defined near a except possibly at a . If f ( x ) ≤ g ( x ) ≤ h ( x ) for x close to a , but not a , lim x → a f ( x ) = L , lim x → a h ( x ) = L , then lim x → a g ( x ) = L . Come up with a version of the theorem where the conclusion is x → a g ( x ) = ∞ . lim Jeffrey Im MAT137 October 10, 2018 6 / 16
Infinite version of the squeeze theorem Theorem Let a ∈ R and let f and g be functions defined near a , except possibly at a . If for x close to a but not a , f ( x ) ≤ g ( x ), and lim x → a f ( x ) = ∞ , then x → a g ( x ) = ∞ . lim Now let’s prove it. 1. Make the hypothesis more precise. 2. Write down what you want to prove. 3. Do some scratch work. 4. Write down a proof. Jeffrey Im MAT137 October 10, 2018 7 / 16
Infinite version of the squeeze theorem Proof. Let M ∈ R be given. From our hypotheses, we know that 1. there exists a δ > 0 such that 0 < | x − a | < δ implies f ( x ) > M , 2. and f ( x ) ≤ g ( x ). Using these hypotheses together, we have that g ( x ) ≥ f ( x ) > M whenever 0 < | x − a | < δ . This shows that lim x → a g ( x ) = ∞ , as desired. Jeffrey Im MAT137 October 10, 2018 8 / 16
A composition theorem Write a formal proof for Theorem Let a , L ∈ R . Let f and g be functions. IF x → a f ( x ) = L lim g is continuous at L THEN lim x → a g ( f ( x )) = g ( L ) Jeffrey Im MAT137 October 10, 2018 9 / 16
A composition theorem Proof. Let ε > 0 be given. Our hypotheses guarantee these two points: 1. (continuity of g at L ) there exists an η > 0 such that | y − L | < η implies | g ( y ) − g ( L ) | < ε . 2. (lim x → a f ( x ) = L ) there exists a δ > 0 such that 0 < | x − a | < δ implies | f ( x ) − L | < η . In point (2), we chose δ so that 0 < | x − a | < δ implies | f ( x ) − L | < η and in point (1), we chose η so that this guarantees | g ( f ( x )) − g ( L ) | < ε . This shows that lim x → a g ( f ( x )) = g ( L ). Jeffrey Im MAT137 October 10, 2018 10 / 16
Intermission Exercise Show that x → 0 ( x 3 + x 2 ) = 0 . lim First proof we came up with today Let ε > 0 be given. Then for | x | < 1 / 2 we have that | x 2 + x | < 1. Pick δ := min { ε, 1 / 2 } . Then 0 < | x | < δ implies | x 3 + x 2 | = | x || x 2 + x | < | x | · 1 < ε. 2nd one � � 3 Let ε > 0 be given and pick δ := min { ε/ 2 , ε/ 2 } . Then 0 < | x | < δ implies | x 3 + x 2 | ≤ | x | 3 + | x | 2 < ε/ 2 + ε/ 2 = ε. Jeffrey Im MAT137 October 10, 2018 11 / 16
New continuous functions We want to prove the following theorem Theorem IF f and g are continuous functions THEN h ( x ) = max { f ( x ) , g ( x ) } is also a continuous function. You are allowed to use all results that we already know. What is the fastest way to prove this? Hint: What is the number a + b + | a − b | ? 2 There is a way to prove this quickly without writing any epsilons. Jeffrey Im MAT137 October 10, 2018 12 / 16
New continuous functions Proof. 1. Show that a + b + | a − b | = max { a , b } . (Cases... a ≥ b and a < b ). 2 2. So max { f ( x ) , g ( x ) } = f ( x )+ g ( x )+ | f ( x ) − g ( x ) | . 2 3. ( f ( x ) + g ( x )) is continuous. 4. ( f ( x ) − g ( x )) is continuous and so | f ( x ) − g ( x ) | is also continuous. 5. From the previous two points, f ( x ) + g ( x ) + | f ( x ) − g ( x ) | is continuous. 6. Lastly, f ( x ) + g ( x ) + | f ( x ) − g ( x ) | = max { f ( x ) , g ( x ) } 2 is continuous, as needed. Jeffrey Im MAT137 October 10, 2018 13 / 16
A question from last year’s test The only thing we know about the function g is that g ( x ) lim = 2 . x 2 x → 0 Use it to compute the following limits (or explain that they do not exist): g ( x ) 1. lim x x → 0 g ( x ) 2. lim x 4 x → 0 g (3 x ) 3. lim x 2 x → 0 Jeffrey Im MAT137 October 10, 2018 14 / 16
Computations! sin x Using that lim = 1, compute the following limits: x x → 0 sin e x sin x 1. lim 4. lim e x x x → 2 x → 0 1 − cos x sin(5 x ) 5. lim 2. lim x x → 0 x x → 0 tan 10 (2 x 20 ) tan 2 (2 x 2 ) 6. lim 3. lim sin 200 (3 x ) x 4 x → 0 x → 0 Jeffrey Im MAT137 October 10, 2018 15 / 16
Limits at infinity Compute: x 7 − 2 x 5 + 11 x 2 + 2 x + 3 � � 1. lim 4. lim x →∞ 3 x 2 + 4 x + 5 x →∞ x 2 − � � � x 5 + 1 2. lim √ x →∞ x 3 + 2 x 6 + 1 x 2 + 11 √ 5. lim 3. lim 2 x 3 + x 5 + 1 x →∞ x + 1 x →∞ Jeffrey Im MAT137 October 10, 2018 16 / 16
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