Solution Sheet 1.4 Questions 26-31 26. Using the Limit Rules evaluate i) ii) iii) 3 2 +4+1 0 2 +4+3, 3 2 +4+1 2 +4+3, 3 2 +4+1 1 2 +4+3. 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 3 2 +4+1 2 +4+3 is well-defined at 0 (in particular the denominator is not 0) so by the Quotient Rule for its 3 2 +4+1 0 2 +4+3 = 0(3 2 +4+1) 0 ( 2 +4+3) = 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 3 2 +4+1 + 2 +4+3 3+4/+1/ 2 = + 1+4/+3/ 2 = + (3+4/+1/ 2 ) + (1+4/+3/ 2 ) by the Quotient Rule = 3 1 = 3. 37
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 2 +4+3 = (+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 3 2 +4+1 = (+1)(3+1). 3 2 +4+1 1 2 +4+3 (+1)(3+1) = 1 (+1)(+3) 3+1 = 1 +3. WecannowapplytheQuotientRuleforitsforboth 1 (3+1) and 1 (+3) eist and the second one is non-zero. Hence 3 2 +4+1 1 2 +4+3 = 1(3+1) 1 (+3) = 2 2 = 1. 27. (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
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 ) 3. 28. 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 1. 0 2 39
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 2 2 = 1 ( ) e 1 + 1 ( ) e 1. 2 2e )+ e 2 ( ) e 1 Now use the Sum and Product Rules for its to get sinh = 1 ( ) ( ) e 0 2 1 1 e + 0 2 0 e 1 0 = 1 2 + 1 2 = 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, 2 2 2 1 as 0. Becausethiseistsandisnon-zerowecanapplytheQuotient Rule to get tanh = 0 sinh 0 0 cosh = 1 1 = 1. 40
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, 1 2 1 2 as 0, by Part i. Thus cosh 1 = 1 0 2 2. 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
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
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
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 2 +1. 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
Finally, we can use the results just found to plot the graph of tanh : 1 0.5-6 -4-2 0 2 4 6-0.5-1 31. i. Prove that e 1 2 2 3 6 < 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 = 1 0 3 6. 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 1 2 2 3 3! = N N k=4 k N 4 k! = j 4 N (j +4)!. (10) j=0 45
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 3 1 1 1 < 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 1 2 2 3 3! 2 4! 4. ii) Divide through the result of part i by 3 to get e 1 2 /2 1 3 6 < 2 < (11) 4! for < 1/2. 46
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 /2 1 3 6 < < δ ε. Hence we have verified the ε-δ definition of e 1 2 /2 = 1 0 3 6. (12) Alternatively we can use the Sandwich Rule for (11) opens out as 1 6 < e 1 2 /2 < 1 3 6 +. 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 2 2 3. 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) 2 3 2 3 = (e 1 2 /2) 2 3 + (e 1 ( ) ( ) 2 /2) 2( ) 3. 47
Let 0 (in which case 0) when, by the assumption of the question, we get sinh 0 3 = 1 2 (e 1 2 /2) 0 3 + 1 2 (e 1 ( ) ( ) 2 /2) 0 2( ) 3 = 1 2 1 6 + 1 2 1 6 = 1 6. Again, the graph of (sinh )/ 3 is not particularly eciting : 0.176 0.174 0.172 0.170 0.168 1.0 0.5 0.0 0.5 1.0 48