Solution Sheet 1.4 Questions 26-31

Transcription

1 Solution Sheet 1.4 Questions Using the Limit Rules evaluate i) ii) iii) , , Note When using a Limit Rule you must write down which Rule you are using and you must show that any necessary conditions of that rule are satisfied. Solution i) The rational function is well-defined at 0 (in particular the denominator is not 0) so by the Quotient Rule for its = 0( ) 0 ( ) = 1 3. ii) We cannot apply the Quotient Rule for its directly since the polynomials on the numerator and denominator diverge as +. Instead, divide top and bottom by the largest power of to get /+1/ 2 = + 1+4/+3/ 2 = + (3+4/+1/ 2 ) + (1+4/+3/ 2 ) by the Quotient Rule = 3 1 = 3. 37

2 iii) We cannot apply the Quotient Rule for its since the denominator is 0 at = 1. This means that the denominator has a factor of +1 and in fact = (+1)(+3). For the it of the rational function to eist the numerator will also have to be zero at = 1, i.e. have a factor of +1. In fact Thus = (+1)(3+1) (+1)(3+1) = 1 (+1)(+3) 3+1 = WecannowapplytheQuotientRuleforitsforboth 1 (3+1) and 1 (+3) eist and the second one is non-zero. Hence = 1(3+1) 1 (+3) = 2 2 = (i) What is wrong with the argument: π ) π ) sin( 0 3 = 3 sin( 0 0 by the Product Rule π ) = 0 sin( 0 = 0. (ii) Evaluate π ) sin( 0 3. Solution i) You may only use the Product Rule for its when both individual its eist. Here we know from an earlier question that 38

3 0 sin(π/) does not eist, so we cannot apply the Product Rule (even if the answer it gives is correct!) ii) We might guess that the it is 0. Let ε > 0 be given. Choose δ = ε 1/3. Then for : 0 < 0 < δ we have ( π ( 3 π ) sin 0 = ) 3 sin 3 since sin(π/) 1, Hence we have verified the definition of π ) sin( 0 3 = 0. = 3 < δ 3 since 0 < δ < ( ε 1/3) 3 = ε since δ = ε 1/3. Alternatively you could use the Sandwich Rule on ( π 3 3 sin ) Recall that in the lectures we have shown that e 1 0 e = 1 and = 1. 0 Use these to evaluate the following its which include the hyperbolic functions. (i) ii) iii) sinh 0, 0 tanh, cosh

4 Solution i) Start from sinh = e e. 2 The guiding principle is to manipulate this so we see a function whose it we already know. For eample (e 1)/. For this reason we add in zero in the form 0 = 1+1 : sinh = e 1+1 e = 1 ( e = 1 ( ) e ( ) e e )+ e 2 ( ) e 1 Now use the Sum and Product Rules for its to get sinh = 1 ( ) ( ) e e e 1 0 = = 1. ii) With the intention of using known results write tanh = sinh 1 cosh. Before we apply the Quotient Rule for its we need to note that cosh = e +e = 1 (e + 1e ) 1 ( 1+ 1 ) = 1, as 0. Becausethiseistsandisnon-zerowecanapplytheQuotient Rule to get tanh = 0 sinh 0 0 cosh = 1 1 = 1. 40

5 We have used Part i in the numerator. iii) Apply the same idea of multiplying by 1 as used for (cos 1)/ 2 in lectures: For 0, cosh 1 2 = cosh 1 2 ( ) cosh+1 = cosh2 1 cosh+1 2 (cosh+1) = ( sinh ) 2 1 cosh+1 since cosh 2 sinh 2 = 1, as 0, by Part i. Thus cosh 1 = The graphs of these functions are not particularly interesting, but I have plotted the graph of sinh/ in black, tanh/ in blue and of (cosh 1)/ 2 in red: y 1 1/2 29. Recall that in the lectures we have shown that sinθ θ 0 θ = 1 Use this to evaluate (without using L Hôpital s Rule) 41

6 i) ii) θ 0 θ tanθ, sinθ tanθ θ 0 θ 3. Solution i) Again guided by the its we already know write θ 0 θ tanθ = θcosθ θ 0 sinθ = θ 0 cosθ ( sinθ θ ) = θ 0cosθ θ 0 sinθ θ by Quotient Rule for its, allowable since both its eist and the it on the denominator is non-zero. Hence Graphically θ 0 y θ tanθ = 1 1 = 1., 1 2π π π/2 π/2 π 2π ii) The it we already know from lectures is of (cosθ 1)/θ 2 so write sinθ tanθ θ 3 = tanθ ( ) cosθ 1 θ θ 2. 42

7 The trick used in lectures to evaluate the it of this it is to multiply top and bottom by cosθ+1 to get ( ) ( ) tanθ cos 2 2 θ 1 1 sinθ θ cosθ+1 = tanθ 1 θ θ cosθ+1 θ 2 = 1 cosθ ( sinθ Use the Product and Quotient Rules for its to deduce Graphically sinθ tanθ θ 0 θ 3 = 1 2. y θ ) 3 1 cosθ+1. π/2 π/2 1/2 30. i) Assuming that e > for all > 0 verify the ε-x definitions of + e = 0 and e = 0. Deduce (using the Limit Rules) that tanh = 1 and + tanh = 1. Sketch the graph of tanh. Solution i) Let ε > 0 be given. Choose X = 1/ε > 0. Assume > X. By the assumption in the question we have e > so 0 < e = 1 e < 1 < 1 X = 1 (1/ε) = ε. 43

8 Thus we have verified the ε-x definition of + e = 0. Let ε > 0 be given. Choose X = 1/ε < 0. Assume < X. This means that is negative, so can be written as = y where y > X = 1/ε. Then, as above, e = e y < 1 y < 1 ( X) = 1 (1/ε) = ε. Thus we have verified the ε-x definition of e = 0. ii) By definition tanh = sinh cosh = e e e +e. For + divide top and bottom by e so tanh = 1 e 2 1+e 2. By the Product Rule for its, part i of this question gives ( + e 2 = ) ( ) 2 e 2 = + + e = 0. Then, by the Quotient Rule for its, tanh = + (1 e 2 ) + + (1+e 2 ) = 1. For divide top and bottom by e so tanh = e2 1 e Again the Product Rule for its and part i gives e2 = 0. Then, by the Quotient Rule for its, tanh = + (e 2 1) + + (e 2 +1) = 1. 44

9 Finally, we can use the results just found to plot the graph of tanh : i. Prove that e < 2 4 4! for < 1/2. Hint Use the method seen in the notes where it was shown that e 1 < 2 for < 1/2. ii. Deduce e 1 2 /2 = iii. Use Part ii. to evaluate sinh. 0 3 Solution i) Start from the definition of an infinite series as the it of the sequence of partial sums, so e ! = N N k=4 k N 4 k! = j 4 N (j +4)!. (10) j=0 45

10 Then, by the triangle inequality, (applicable since we have a finite sum), N 4 j N 4 j (j +4)! (j +4)! 1 N 4 j 4! j=0 j=0 = 1 4! ( j=0 since (j +4)! 4! for all j 0, ) 1 N 3, 1 on summing the Geometric Series, allowable when = 1. In fact we have < 1/2 < 1, which means 1 N < 1 1 1/2 = 2. Hence N 3 j=0 j (j +4)! 2 4! for all N 0. Therefore, since the it of these partial sums eists the it must satisfy N 3 j N (j +4)! 2 4!. j=0 Combined with (10) we have e ! 2 4! 4. ii) Divide through the result of part i by 3 to get e 1 2 / < 2 < (11) 4! for < 1/2. 46

11 To prove the it in the question we can verify the definition. Let ε > 0 be given. Choose δ = min(1/2,ε). Assume 0 < 0 < δ. Since δ 1/2, the inequality (11) holds for such. Thus e 1 2 / < < δ ε. Hence we have verified the ε-δ definition of e 1 2 /2 = (12) Alternatively we can use the Sandwich Rule for (11) opens out as 1 6 < e 1 2 /2 < Let 0 when the upper and lower bound 1/6. Thus, by the Sandwich Rule, (12) follows. iii) From the definition of sinh we have sinh 3 = e e This has to be manipulated so that we see e 1 2 /2 and can thus use (12). Do this by adding in zero in the form to get 0 = 2 /2 ( ( ) 2 2 ), e e 2 = (e 1 2 /2) (e 1 ( ) ( ) 2 /2) = (e 1 2 /2) (e 1 ( ) ( ) 2 /2) 2( ) 3. 47

12 Let 0 (in which case 0) when, by the assumption of the question, we get sinh 0 3 = 1 2 (e 1 2 /2) (e 1 ( ) ( ) 2 /2) 0 2( ) 3 = = 1 6. Again, the graph of (sinh )/ 3 is not particularly eciting :