Limits PDF

CHAPTER 05 Limits Learning Part Session 1 Definition of Limits Indeterminate Forms L’Hospital’s Rule Evaluation of Limits Session 2 Trigonometric Limits Session 3 Logarithmic Limits Exponential Limits Session 4 Miscellaneous Forms Session 5 Left Hand and Right Hand Limits Session...

CHAPTER 05 Limits Learning Part Session 1 Definition of Limits Indeterminate Forms L’Hospital’s Rule Evaluation of Limits Session 2 Trigonometric Limits Session 3 Logarithmic Limits Exponential Limits Session 4 Miscellaneous Forms Session 5 Left Hand and Right Hand Limits Session 6 Use of Standard Theorems/Results Use of Newton-Leibnitz’s Formula in Evaluating the Limits Summation of Series Using Definite Integral as the Limit Practice Part JEE Type Examples Chapter Exercises Arihant on Your Mobile ! Exercises with this #L symbol can be practised on your mobile. See title inside to activate for free. Session 1 Definition of Limits, Indeterminate Forms, L’Hospital’s Rule, Evaluation of Limits Definition of Limits (b) When the left tendency is not the same as its right tendency Here, left tendency is l1 and right Let lim f ( x ) = l. It would mean that when we approach x ®a tendency is l 2 , clearly left tendency (l 1 ) is not same as the point x = a from the values which are just greater than right tendency (l 2 ). In this case, we say that the limit of or smaller than x = a, f ( x ) would have a tendency to move f ( x ) at x = a will not exist. closer to the value ‘l ’. i.e. lim f ( x ) = Doesn’t exist. This is same as saying ‘difference between f ( x ) and l can x ®a be made as small as we feel like by suitably choosing x in Y the neighbourhood of x = a’. Mathematically, we write this, as lim f ( x ) = l, which is y = l2 x ®a equivalent of saying that, | f ( x ) – l | < e , " x whenever y = l1 0 < | x - a | < d and e and d sufficiently small +ve numbers. X O x=a It is clear from the above discussion that, if we are x=a+h interested in finding the limit of f ( x ) at x = a, the first x=a–h thing we have to make sure that f ( x ) is well defined in the neighbourhood of x = a and not necessarily at x = a Figure 5.2 (that means x = a may or may not be in the domain of f ( x )), because we have to examine its behaviour or (c) When the left tendency and/or right tendency is tendency in the neighbourhood of x = a. not fixed As shown in the figure 5.3, it is clear that in this case, the function has erratic behaviour in the Following possibilities may arise : neighbourhood of x = a and it will not be possible to (a) Left tendency is same as its right tendency As talk about the left and right tendencies of the function shown in figure 5.1, when we approach x = a from in the neighbourhood of x = a. the values which are just less than a, f ( x ) has a In this case, we conclude that the limit of f ( x ) at tendency to move towards the value l (left tendency). x = a will not exist. Similarly, when we approach x = a from the values i.e. lim f ( x ) = Doesn’t exist. which are just greater than a, f ( x ) has a tendency to x ®a move towards the value l (right tendency). Y In this case, we say f ( x ) has limit l at x = a, i.e. lim f ( x ) = l. x ®a Y y = f(x) y = f(x) y=l O X x a O x=a Figure 5.3 Figure 5.1  Chap 05 Limits 247 Remarks Infinity ( ¥) is a symbol and not a number. It is a symbol 1. Normally, students have the Y for the behaviour of a variable which continuously perception that limit should be a increases and passes through all limits. Thus, the finite number. But, it is not always so. It is quite possible that f ( x ) has statement x = ¥ is meaningless, we should write x ® ¥. infinite limit at x = a. Similarly, – ¥ is a symbol for the behaviour of a variable If xlim ® a f ( x ) = ¥, it would simply O X which continuously decreases and passes through all mean that function has tendency to limits. Thus, the statement x = – ¥ is meaningless, we assume very large positive values in Figure 5.4 neighbourhood of x = a (as shown in should write x ® – ¥. Y figure 5.4). 1 1 1 Also, ® 0, if x ® + ¥ and ® 0, if x ® – ¥. For example lim =¥ O X x x x ® 0 | x| which indicates the left tendency as We cannot plot ¥ on paper. Infinity does not obey laws of well as right tendency are the same. elementary algebra. Again, if lim f ( x ) = - ¥, it would (i) ¥ + ¥ = ¥ is indeterminate x ® a simply mean that the function has Figure 5.5 tendency to assume very large (ii) ¥ – ¥ is indeterminate. negative values in the neighbourhood of x = a as shown in figure 5.5. For example (as shown in figure 5.5) lim –1 =– ¥ L’Hospital’s Rule x ® 0 |x| f (x ) 0 ¥ At the end we discuss the case when left tendency is (– ¥) and This rule states that, if lim , reduces to or × right tendency is ( + ¥) (i.e. f ( x ) does not have unique x ®a g( x ) 0 ¥ tendency). Then, differentiate numerator and denominator till this Thus, in this case limit does not exist. form is removed. 1 Y For example lim does not exist, f (x ) f ¢ (x ) x ® 0 x i.e. lim = lim , provided the later limit exists. since left tendency is (– ¥) and right x ® a g( x ) x ® a g ¢ (x ) tendency is ( + ¥) as shown in X figure 5.6. O æ 0 ¥ö But, if it again take form ç or ÷ , 2. If f ( x ) is well defined at x = a, it è 0 ¥ø doesn’t imply that Figure 5.6 f (x ) f ¢ (x ) f ¢¢( x ) lim f ( x ) = f ( a). Because, it is quite then lim = lim = lim x ® a possible that f ( x ) is well x ®a g ( x ) x ® a g ¢ ( x ) x ® a g ¢¢( x ) defined at x = a but not in the neighbourhood of x = a or f ( x ) is æ0 ¥ö well defined in the neighbourhood of x = a, but doesn’t have a and this process is continued till ç or ÷ form is removed. è0 ¥ø unique tendency. Remark Indeterminate Forms L’Hospital’s rule is applicable to only two indeterminate ¥ forms æç or ö÷. 0 If direct substitution of x = a while evaluating lim ( x ) è0 ¥ø x ®a leads to one of the following forms 0 ¥ x 6 – 24 x – 16 , , ¥ – ¥, 1 ¥ , 0 0 , ¥ 0 , ¥ ´ 0 then it is called e.g. Evaluate lim × 0 ¥ x ®2 x 3 + 2x – 12 indeterminate form. x 6 – 24 x – 16 é0 ù Sol. We have, lim êë 0 formúû x2 –1 0 x ®2 x + 2x – 12 3 e.g. lim = indeterminate form. x ®1 x – 1 0 6x 5 – 24 = lim [by L’Hospital’s rule] x n – an 0 x ®2 3x 2 + 2 lim = indeterminate form. x ®a x – a 0 6 (2)5 – 24 168 = = = 12 sin x 0 3 ( 2) + 2 2 14 lim = indeterminate form. x ®0 x 0 248 Textbook of Differential Calculus é 23 x 3 25 x 5 ù é x3 x5 ù Frequently Used Series Expansions x ê2x + 3 +2 15 +...ú –2x ê x + 3 +2 15 +...ú 1. ex = 1 + x + x2 x3 + +... = lim ë û ë û 1! 2 ! 3! x ®0 (2 sin 2 x )2 x × log a (log a) 2 x 2 æ8 2ö æ 64 4ö 2. ax = 1 + + +... x 4 ç - ÷ + x 6 ç - ÷ +... 1! 2! è3 3ø è 15 15 ø = lim [where, a Î R+ ] x ®0 æ x3 x5 ö 4 nx n ( n – 1) x 2 4 çx - + –...÷ 3. ( 1 + x ) n = 1 + + è 3! 5! ø 1! 2! 2 + 4 x 2 +... 2 1 + n ( n – 1) ( n – 2) x 3 +... n Î R and|x| < 1 = lim 4 = = x ®0 4 2 3! æ x 2 ö 4 ç1 - +...÷ 4. log ( 1 + x ) = x – x 2 + x – x 3 +... 4 è 3 ! ø 2 3 4 [where –1 £ x £ 1] 5. x n – an x–a = x n – 1 + x n – 2a + x n – 3a2 +... + an – 1 Evaluation of Limits Now, according to our plan, first of all we shall learn the æ x 11x 2 ö 6. ( 1 + x )1 /x = e çç1 – + +...÷÷ evaluation of limits of different forms and then learn the è 2 24 ø existence of limits. x 3 x5 x 7 There are eight indeterminate or meaningless forms, 7. sin x = x – + – +... 3! 5 ! 7! which are x 2 x4 x6 0 ¥ 8. cos x = 1 – + – +... (i) (ii) (iii) ¥ – ¥ (iv) ¥ ´ ¥ 2! 4 ! 6 ! 0 ¥ x3 2 5 17 7 (v) ¥ × 0 (vi) 0 0 (vii) ¥ 0 (viii) 1 ¥ 9. tan x = x + + x + x +... 3 15 315 We will divide the problems of evaluation of limits in five 12 3 12 × 32 5 12 × 32 × 52 7 categories, which are 10. sin–1 x = x + x + x + x +... 3! 5! 7! 1. Limit of algebraic functions x 3 x5 2. Trigonometric limits 11. tan–1 x = x – + +... 3 5 3. Logarithmic limits x2 5x 4 61x 6 12. sec –1 x = 1 + + + +... 4. Exponential limits 2! 4! 6! 5. Miscellaneous forms 2 2 2 × 22 4 2 × 22 × 4 2 6 13. (sin–1 x ) 2 = x + x + x +... Now, we discuss one by one in details. 2! 4! 6! x 3 x 4 2x 6 14. x cot x = 1 – 3 + – 45 945 +... Limit of Algebric Functions 2 4 6 15. sec x = 1 + x + 5x + 61x +... In this section, we evaluate limit of algebraic functions 2 24 720 when variable tends to a finite or infinite value. While x2 7x4 31x 6 0 ¥ 16. x cosec x = 1 + + + +... evaluating algebraic limits the form , and ¥ - ¥ arise, 6 360 15120 0 ¥ which we will discussed here. x tan 2x – 2x tan x 0 y Example 1 Evaluate lim 2 × (i) Form x ®0 (1 – cos 2x ) 0 [IIT JEE 1999] This form can be resolved by factorisation method, x tan 2x – 2x tan x rationalisation method or by using the formula Sol. We have, lim x ®0 (1 – cos 2x )2 x n - an lim = na n - 1 , which are discussed below. x ®a x - a  Chap 05 Limits 249 (a) Factorisation Method x+h– x y Example 4 Evaluate lim. In this method, numerators and denominators are h ®0 h factorised. The common factors are cancelled and the rest x +h – x output is the result. Sol. Method I We have, lim h ®0 h x 2 - 3x + 2 x +h – x x +h + x é0 ù y Example 2 Evaluate lim × = lim ´ êë 0 form úû x ®1 x –1 h ®0 h x +h + x x 2 – 3x + 2 ( x + h )–( x ) h é0 ù = lim = lim Sol. Method I We have, lim êë 0 form úû h ®0 h( x + h + x ) h ® 0 h( x + h + x ) x ®1 x –1 ( x – 1)( x – 2) 1 1 = lim = lim = x ®1 ( x – 1) h ®0 x +h + x 2 x [as x 2 - 3x + 2 = ( x - 1)( x - 2)] Method II (L’Hospital’s rule) We have, = lim ( x – 2) [as x –1 ¹ 0] x +h – x é0 ù x ®1 L = lim êë 0 form úû h ®0 h = 1–2 = - 1 \ Applying L’Hospital’s rule, x 2 – 3x + 2 é0 ù 1 Method II We have, L = lim êë 0 form úû –0 x ®1 x –1 2 x +h L = lim So, applying L’Hospital’s rule, h ®0 1 2x – 3 2 – 3 [differentiating numerator and denominator w.r.t. h] L = lim = =–1 x ®1 1 1 1 = [i.e. differentiating numerator and 2 x denominator separately] x - 2a + x - 2a y Example 5 Evaluate lim. x 3 – x 2 log x + log x – 1 x ® 2a y Example 3 Evaluate lim. x 2 - 4a 2 x ®1 x 2 –1 x - 2a + x - 2a é0 ù Sol. We have, lim êë 0 form úû x 3 – x 2 log x + log x – 1 x ® 2a x - 4a 2 2 Sol. We have, lim x ®1 x2 –1 x - 2a x - 2a + ( x 3 – 1)–( x 2 – 1) log x é0 ù x + 2a = lim = lim x ®1 ( x 2 – 1) êë 0 form úû x ® 2a x - 2a x + 2a 1 x - 2a ( x –1){ x 2 + x + 1 – ( x + 1) log x } = lim + = lim x ® 2a x + 2a x + 2a ( x + 2a ) x ®1 ( x –1)( x + 1) éQ x 3 - 1 = ( x - 1) ( x 2 + x + 1),ù 1 1 = = ê ú 4a 2 a êë x - 1 = ( x - 1) ( x + 1) 2 úû x 2 + x + 1 – ( x + 1) log x (c) Based on Standard Formula = lim x ®1 ( x + 1) x n – an lim = na n – 1 , where n is a rational number. 12 + 1 + 1 – (1 + 1) log 1 3 x ®a x –a = = [as log 1 = 0] 1+1 2 x n – an Proof Let f ( x ) = x –a (b) Rationalisation Method = x n – 1 + ax n – 2 + a 2 x n – 3 +...+ a n – 1 Rationalisation is followed when we have fractional 1 1 \ lim f ( x ) = lim ( x n – 1 + ax n – 2 + a 2 x n – 3 +...+ a n – 1 ) x ®a x ®a powers (like , etc.) on expressions in numerator or 2 3 =a + a × a n – 2 + a 2 × a n – 3 +...+ a n – 1 n–1 denominator or in both. After rationalisation the terms are 1444444 424444444 3 factorised which on cancellation gives the result. n terms = a n – 1 + a n – 1 + a n – 1 + K upto n terms = n × a n – 1 250 Textbook of Differential Calculus f (x) = ax, when 0 1. This graph appears ì x - 1 x 2 - 12 x 3 - 13 x n - 1n ü to touch X-axis in the negative side of X-axis and = lim í + + +...+ ý x ®1 x - 1 x -1 x -1 x -1 þ thereafter it increases rapidly. This is why because î lim a x ® 0, again you will also find the result, = 1 + 2 (1)2 -1 + 3 (1)3 -1 +... + n (1)n -1 x ®-¥ lim a x ® ¥ = 1 + 2 + 3 +... + n x ®¥ = n ( n + 1) ì¥, if a > 1 ï 2 Thus, we have lim a = í1, if a = 1 x x ®¥ 3 ï0, if 0 £ a < 1 y î y Example 8 The value of lim as x ®1 x3 - y 2 -1 This type of problems are solved by taking the highest y ®0 power of the terms tending to infinity as common from ( x , y ) ® (1, 0) along the line y = x - 1 is numerator and denominator. That is after they are (a) 1 cancelled and the rest output is the result (or apply (b) -1 L’Hospital’s rule). (c) 0 x2 + 5 (d) Doesn’t exist y Example 9 Evaluate lim. x ®¥ x 2 + 4x + 3 Sol. As, y ® x - 1 or x ® y + 1 x2 + 5 Sol. Let L = lim (y )3 x ®¥ x + 4x + 3 2 \ lim y ®0 ( y + 1) - y - 1 3 2 Dividing numerator and denominator by x 2 , we get 5 Using L, Hospital’s rule, 1+ 2 x 1+0 3y 2 6y L = lim = =1 lim = lim x ®¥ 4 1+ + 2 3 1 + 0 + 0 y ®0 3 (y + 1) - 2y 2 y ®0 6 ( y + 1) - 2 x x 0 K =0 = [because ® 0, when x ® ¥, where K is any constant] 6 x Hence, (c) is the correct answer. x2 + 5 é¥ ù Aliter We have, L = lim êë ¥ form úû x ®¥ x + 4x + 3 2 (ii) Algebraic Function of ¥ Type 2x é¥ ù ¥ Applying L’Hospital’s rule, L = lim ® ¥ 2x + 4 êë ¥ form úû (a) Form x ¥ 2 Again, applying L’Hospital’s rule, L = lim =1 First we should know the limiting values of a x (a > 0 ) as x ® ¥2 x ® ¥. See the graphs of this function.  Chap 05 Limits 251 (n + 2)! + (n + 1)! An Important Result y Example 10 Evaluate lim. n ®¥ (n + 2)! – (n + 1)! If m, n are positive integers and a0, b0 ¹ 0 and non-zero real (n + 2)(n + 1) ! + (n + 1) ! numbers, then Sol. We have, lim a x m + a1x m – 1 + K + am – 1x + am n ®¥ (n + 2)(n + 1) ! – (n + 1) ! lim 0 n x ® ¥ b x + b xn – 1 + K + b n – 1x + bn (n + 1) ![ n + 2 + 1] ( n + 3) é¥ ù 0 1 = lim = lim êë ¥ form úû ì 0, m n when a0b0 > 0 ï ïî– ¥, m > n when a0b0 < 0 (b) ¥ - ¥ Form Such problems are simplified (generally rationalised) first, ax 2 + b y Example 13 Evaluate lim , when a ³ 0. æ ¥ö x ®¥ x + 1 thereafter they generally acquire ç ÷ form. è ¥ø ax 2 + b Sol. Here, if a ¹ 0 ; lim y Example 11 Evaluate lim ( x – x 2 + x ). x ®¥ x + 1 x ®¥ [as degree of numerator > degree of denominator] Sol. We have, lim ( x – x 2 + x ) ax 2 + b x ®¥ = lim =¥ [as a > 0] x ®¥ x + 1 x – x2 + x x + x2 + x 0× x 2 + b = lim ´ [ ¥ – ¥ form] Again, if a = 0 ; lim = lim b =0 x ®¥ 1 x + x2 + x x ®¥ x +1 x ®¥ x + 1 x 2 – (x 2 + x ) –x [as degree of numerator < degree of denominator] = lim = lim ax 2 + b ì ¥, a > 0 x ®¥ x+ x +x 2 x ®¥ x + x2 + x \ lim =í x ®¥ x + 1 î0, a = 0 –x = lim æx2 +1 ö x ®¥ ì 1ü y Example 14 If lim ç – ax – b ÷ = 0, find the x í1 + 1 + ý x ®¥ è x + 1 î xþ ø values of a and b. –1 1 é 1 ù æx2 + 1 ö = lim =– êëQ x ® 0, as x ® ¥ úû x ®¥ 1 2 Sol. Given, lim ç – ax – b ÷ = 0 1+ 1+ x ®¥ è x + 1 ø x x 2 + 1 – ax 2 – ax – bx – b Þ lim =0 y Example 12 Evaluate lim ( x + x + 1 – x + 1 ). 2 2 x ®¥ x +1 x ®¥ x 2 (1 – a ) – x (a + b ) + (1 – b ) Sol. We have, lim ( x 2 + x + 1 – x 2 + 1 ) [ ¥ – ¥ form ] Þ lim =0 x ®¥ x ®¥ x +1 Since, the limit of above expression is zero. ( x 2 + x + 1 – x 2 + 1) ( x 2 + x + 1 + x 2 + 1) = lim ´ \ Degree of numerator < Degree of denominator. x ®¥ 1 ( x 2 + x + 1 + x 2 + 1) So, numerator must be a constant, i.e. a zero degree polynomial. ( x 2 + x + 1) – ( x 2 + 1) = lim \ 1 – a = 0 and a + b = 0 x ®¥ x2 + x + 1 + x2 + 1 Hence, a = 1 and b = –1 x æx2 +1 ö = lim y Example 15 If lim ç – ax – b ÷ = 2, find the x ®¥ x2 + x + 1 + x2 + 1 x ®¥ è x + 1 ø 1 1 1 values of a and b. = lim = = x ®¥ 1 1 1 1+1 2 æx2 + 1 ö 1+ + 2 + 1+ 2 Sol. We have, lim ç – ax – b ÷ x x x x ®¥ è x + 1 ø é 1 ù x 2 (1 – a ) – x (a + b ) + (1 - b ) = lim =2 êëQ x ® 0, as x ® ¥ úû x ®¥ ( x + 1) 252 Textbook of Differential Calculus Since, limit of above expression is a finite non-zero number. n ( n + 1) Sol. As, Sn = \ Degree of numerator = Degree of denominator 2 Þ 1–a = 0 Þ a = 1 n ( n + 1) n 2 + n - 2 ( n + 2) ( n - 1) Þ Sn - 1 = -1= = Putting a = 1 in above limit, we get 2 2 2 - x (1 + b) + (1 + b ) Sn æ n ö æn + 1ö lim =2 \ =ç ÷ç ÷ x ®¥ x +1 Sn - 1 è n - 1 ø è n + 2 ø Þ – (1 + b ) = 2 Þ b = –3 æ2 3 4 n öæ3 4 5 n + 1ö Hence, a = 1 and b = - 3. Þ Pn = ç × × K ÷ç × × K ÷ è1 2 3 n - 1ø è 4 5 6 n + 2ø æx2 +1 ö y Example 16 If lim ç – ax – b ÷ = ¥, find a and æn ö æ 3 ö x ®¥ è x + 1 =ç ÷ç ÷ ø è 1 ø èn + 2ø b. 3n \ lim Pn = lim =3 n ®¥ n ® ¥n + 2 æx2 + 1 ö Sol. Given, lim ç – ax – b ÷ = ¥ x ®¥ è x + 1 ø 1 y Example 18 If lim -1 = 1, x lies in the Þ lim x (1 – a ) – x (a + b ) + (1 – b ) 2 =¥ n ®¥ (sin x )n + 1 x ®¥ x +1 interval The limit of above expression is infinity. (a) ( - sin 1, sin 1) (b) ( - 1, 1) \ Degree of numerator > Degree of denominator (c) (0, 1) (d) ( - 1, 0) Þ 1–a > 0 Þa ¹ 1 1 Sol. Here, lim -1 = 1 is possible only, if Hence, a < 1 and b can assume any real value. n ®¥ (sin x )n + 1 y Example 17 Let S n = 1 + 2 + 3 + L + n -1 < sin -1 x < 1 S2 S S S Þ x Î ( - sin 1, sin 1) and Pn = × 3 × 4 K n , Hence, (a) is the correct answer. S 2 - 1 S 3 - 1 S 4 - 1 Sn - 1 where n Î N (n ³ 2). Find lim Pn. n ®¥ Exercise for Session 1 g( x ) f (a ) - g(a ) f ( x ) 1. If f (a ) = 2, f ¢ (a ) = 1, g(a ) = - 1, g ¢ (a ) = - 2, then lim is x ®a x -a (a) - 5 (b) 3 (c) - 3 (d) 5 x cos x - log (1 + x ) 2. The value of lim is x ®0 x2 1 1 (a) 1 (b) (c) (d) None of these 4 2 2 e x - cos x 3. The value of lim is x ®0 x2 3 3 1 1 (a) (b) - (c) (d) - 2 2 2 2 cos x - cos a 4. The value of lim is x ®a cot x - cot a (a) - sin3 a (b) cos3 a (c) sin3 a (d) cot a  Chap 05 Limits 253 æ 1 ö 5. The value of lim ç 2 - cot x ÷ is x ®0 èx ø (a) 0 (b) 1 1 (c) (d) None of these 4 6. The value of lim ( a 2x 2 + ax + 1 - a 2x 2 + 1), (a > 0) is x ®¥ 1 1 (a) (b) - 2 2 (c) Doesn’t exist (d) None of these 13 + 23 + 33 +... + n 3 7. The value of lim is n®¥ (n 2 + 1)2 1 1 (a) (b) 4 2 1 (c) (d) None of these 2 2 1× n + 2 × (n - 1) + 3 × (n - 2) +... + n × 1 8. The value of lim is n®¥ 12 + 22 +... + n 2 (a) 1 (b) -1 1 1 (c) (d) 2 2 an + b n 9. The value of lim , (where a > b > 1) is n ® ¥ an - b n (a) 1 (b) -1 1 1 (c) (d) 2 2 Session 2 Trigonometric Limits Trigonometric Limits y Example 21 Evaluate lim 2 –x sin (2 x ). x ®¥ To evaluate trigonometric limits the following results are 1 given below Sol. Since, 2–x =. We know that, as x ® ¥, 2x ® ¥ 2x sin x tan x (i) lim =1 (ii) lim =1 \ The given limit = 0 ´ x ®0 x x ®0 x [A finite number between –1 and +1] =0 sin–1 x tan–1 x sin (2x ) (iii) lim =1 (iv) lim =1 Hence, lim =0 x ®0 x x ®0 x x ®¥ ( 2x ) sin x 0 p (v) lim = (vi) lim cos x = 1 y Example 22 Evaluate lim e x sin (d /e x ). x ®0 x 180° x ®0 x ®¥ sin ( x – a ) tan ( x – a ) (vii) lim = 1 (viii) lim =1 Sol. When x ® ¥, e ® ¥ x x ®a x –a x ®a x –a d finite But, angle of sine = x = =0 1 – cos x e ¥ y Example 19 Evaluate lim. sin (d/e x ) sin d/e x x ®0 x2 \ The given limit = lim = lim ´d 1–cos x x ®¥ 1/e x d ® 0 d/e x Sol. We have, lim ex x ®0 x2 =1 ´ d = d 2 sin 2 x/ 2 2 sin 2 x/ 2 é0 ù = lim = lim êë 0 formúû x – sin x x ®0 x 2 x ®0 4 x 2/ 4 y Example 23 Evaluate lim. 2 x ®¥ x + cos 2 x 1 æ sin x / 2 ö 1 2 1 = lim ç ÷ = ( 1) = æ sin x ö x ® 0 2 è x /2 ø 2 2 x ç1 – ÷ x – sin x è x ø Sol. We have, lim = lim 1 - cos (1 - cos x ) x ®¥ x + cos 2 x x ®¥ æ cos 2 x ö y Example 20 Solve lim x ç1 + ÷ x ®0 sin 4 x è x ø æ xö sin x 1 - cos ç2sin 2 ÷ 1– 1 - cos (1 - cos x ) è 2ø sin 4 x = lim x = 1–0 =1 Sol. lim = lim x ®0 sin 4 x x ®0 x4 x4 x ®¥ 2 cos x 1+0 1+ x æ xö 1 - cos ç2sin 2 ÷ 4 è 2ø æ sin x ö = lim lim ç ÷ sin 2 x – sin 2 y x ®0 x4 x ®0è x ø y Example 24 Evaluate lim. x ®y x2 – y 2 æ xö 2sin 2 çsin 2 ÷ sin 2 x – sin 2 y sin ( x + y ) sin ( x –y ) è 2ø Sol. lim = lim = lim 1 x ®0 x4 x ®y 2 x –y 2 x ®y ( x + y ) ( x –y ) 2 sin( x + y ) sin( x –y ) sin(2y ) æ æ 2 xö xö = lim ´ lim = ´1 ç sin çsin ÷ sin 2 ÷ x ®y (x +y ) x ® y ( x –y ) 2y è 2 ø = lim 2 × ç × 2÷ x ®0 ç 2 x x 2 ÷ [as x ® y Þ ( x – y ) ® 0, but x + y ® 2y ] ç sin 4× ÷ è 2 4 ø = sin 2y 1 1 2y =2´ = 42 8  Chap 05 Limits 255 y Example 25 \ a is minimum, when x = –1, i.e. a = 2 lim [( x + 5) tan -1 ( x + 5) - ( x + 1) tan -1 ( x + 1)] is equal to Again, b = lim 2 sin 2 q/ 2 = lim 2 sin 2 q/ 2 × 2 x ®¥ q®0 q2 q®04 q /4 (a) p (b) 2p 2 sin 2 q/ 2 1 p = lim × = (c) (d) None of these q ® 0 4 (q/ 2)2 2 2 n n n –r lim (( x + 5) tan -1( x + 5) - ( x + 1) tan -1( x + 1)) æ1ö Sol. Here, x ®¥ Hence, åar bn – r = å 2r × çè 2 ÷ø r =0 r =0 ìp ü = lim ( x + 5) í - cot -1( x + 5)ý - ( x + 1) 1 x ®¥ î2 þ Þ n {20 + 22 + 24 +...+ 22n } 2 ìp -1 ü í - cot ( x + 1)ý 1 ì 1 ( 4n + 1 – 1) ü 4n + 1 – 1 î 2 þ Þ í ý = 2n î 4 –1 þ 2n ´ 3 æ 1 ö -1 æ 1 ö tan -1 ç ÷ tan ç ÷ [i.e. sum of (n + 1) terms of GP] è x + 5ø è x + 1ø = lim 2p - + = 2p x ®¥ æ 1 ö æ 1 ö n 4n + 1 – 1 ç è x + 5ø ÷ ç è x + 1ø ÷ \ åar bn – r = 2n ´ 3 r =0 Hence, (b) is the correct answer. y Example 28 Evaluate sin ( p cos 2 x ) æxö æxö æxö æxö y Example 26 Evaluate lim. lim cos ç ÷ cos ç ÷ cos ç ÷... cos ç n ÷. x ®0 x2 n ®¥ è2ø è4ø è8ø è2 ø [IIT JEE 2001] æxö æ x ö æx ö æx ö æx ö sin ( p cos 2 x ) sin { p (1–sin 2 x )} Sol. Here, lim cos ç n ÷ cos ç n -1 ÷...cos ç ÷ cos ç ÷ cos ç ÷ Sol. lim = lim n ®¥ è2 ø è2 ø è8ø è4ø è2ø x ®0 x2 x ®0 x2 sin ( p – p sin 2 x ) sin 2n A = lim We know, cos A cos 2A cos 22 A...cos 2n -1 A = x ®0 x2 2n sin A ì sin ( p sin 2 x ) p sin 2 x ü æxö æ x ö æx ö æx ö æx ö sin ( p sin 2 x ) Thus, lim cos ç n ÷ cos ç n -1 ÷...cos ç ÷ cos ç ÷ cos ç ÷ = lim = lim í ´ ´ ý n ®¥ è2 ø è2 ø è8ø è4ø è2ø x2 î p sin x 2 x ®0 x ®0 1 x2 þ æxö sin ( p sin 2 x ) sin 2 x sin 2n ç n ÷ = lim ´ p ´ lim è2 ø sin x = lim = lim x ®0 p sin 2 x x ®0 x2 n ®¥ n æxö n ®¥ æxö 2 sin ç n ÷ 2n sin ç n ÷ =1 ´ p ´ 1 = p è2 ø è2 ø 1 x y Example 27 Let a = min { x 2 + 2x + 3, x ÎR} and = sin x × lim × n ®¥ n æxö x 1 – cos q n 2 sin ç n ÷ b = lim. The value of S a r ×b n – r is è2 ø q®0 q 2 r = 0 sin x 1 = × lim ×x Sol. Here, a = min { x 2 + 2x + 3, x Î R } x n ®¥ æxö 2n sin ç n ÷ i.e. x 2 + 2x + 3 = x 2 + 2x + 1 + 2 è2 ø = ( x + 1) 2 + 2 sin x sin x é x q ù = ×1 = n ® ¥, n ® 0 and lim = 1ú x x êë 2 q ® 0 sin q û Exercise for Session 2 1. If lim ( x -3 sin 3x + ax -2 + b ) exists and is equal to zero, then x ®0 9 9 (a) a = - 3, b = (b) a = 3, b = 2 2 -9 (c) a = - 3, b = (d) None of these 2 x sin a - a sin x 2. The value of lim is x ®a x -a (a) a sin a - cos a (b) sin a - a cos a (c) cos a + a sin a (d) sin a + a cos a 2 + cos x - 1 3. The value of lim is x ®p ( p - x )2 1 1 (a) (b) 4 2 (c) 2 (d) Doesn’t exist ( 2 - cos q - sin q) 4. The value of lim is q ® p/ 4 (4q - p )2 1 1 (a) (b) 16 2 16 1 1 (c) (d) 8 2 2 2 (cos x + sin x )3 - 2 2 5. The value of lim is x ® p/ 4 1 - sin 2x 3 3 (a) (b) - 2 2 1 1 (c) (d) - 2 2 Session 3 Logarithmic Limits, Exponential Limits Logarithmic Limits y Example 32 Evaluate lim log e (1 + 2h ) – 2 log e (1 + h ). In this section, we will deal with the problems based on h ®0 h2 expansion of logarithmic series, which is given below [IIT JEE 1999] x2 x 3 loge (1 + 2h ) – 2 loge (1 + h ) log (1 + x ) = x – + +... ¥ Sol. We have, lim h ®0 h2 2 3 é (2h )2 (2h )3 ù æ h2 h3 ö where, –1 £ x £ 1 and it should be noted that the expansion ê(2h ) – + –... ¥ ú –2 çh – + –...÷ is true only if the base is e. To evaluate the logarithmic = lim ë 2 3 û è 2 3 ø log (1 + x ) h ®0 h2 limit, we use lim = 1. x ®0 x –h + 2h –... 2 3 = lim h ®0 h2 log { 1 + ( x – a )} y Example 29 Evaluate lim × h 2 {–1 + 2h –...} x ®a (x – a) = lim h ®0 h2 log {1 + ( x - a )} = lim {–1 + 2h –...} = - 1 Sol. We have, lim h ®0 x ®a ( x - a) ì 1 öü Let x – a = y , when x ® a; y ® 0 y Example 33 Solve lim í x - x 2 × log æç 1 + ÷ ý. x ®¥ î è x øþ log {1 + y } \ The given limit = lim =1 1 ì æ 1 öü y ®0 y Sol. Here, put x = in lim í x - x 2 × log ç1 + ÷ý y x ®¥ î è x øþ log 10 (1 + h ) y Example 30 Evaluate lim × ì 1 log (1 + y )ü = lim í - h ®0 h ý y ®0 y î y2 þ log10 (1 + h ) loge (1 + h ) ´ log10 e Sol. We have, lim = lim y - log(1 + y ) h ®0 h h ®0 h = lim loge (1 + h ) y ®0 y2 = lim ´ log10 e h ®0 h ì y2 y3 y4 ü y - íy - + - + Ký é log(1 + x ) ù î 2 3 4 þ = log10 e ´ 1 = log10 e êëQ xlim = 1ú = lim ®0 x û y ®0 y 2 log ( 5 + x ) – log ( 5 – x ) ì1 y ü y Example 31 Evaluate lim. y 2 í - + Ký x ®0 x = lim î 2 3 þ ì æ x öü ì æ x öü y ®0 y2 log í5 ç1 + ÷ý – log í5 ç1 – ÷ý î è ø 5 þ î è 5 øþ ì1 y ü 1 Sol. We have, lim = lim í - + Ký = x ®0 x y ® 0î2 3 þ 2 æ xö æ xö log 5 + log ç1 + ÷ – log 5– log ç1 – ÷ è 5ø è 5ø Remark = lim x ®0 x If we solve above question as æ xö æ xö ì 1 log ( 1 + y ) ü ì 1 log ( 1 + y ) 1 ü log ç1 + ÷ log ç1 – ÷ lim í - ý = ylim í - × ý è 5ø è 5ø 1 1 2 y®0 îy y2 þ ® 0î y y yþ = lim – = + = x ®0 æx ö æ xö 5 5 5 é log ( 1 + y ) ù 5ç ÷ –5 ç – ÷ êas ylim = 1ú è5ø è 5ø ë ® 0 y û æ1 1ö = lim ç - ÷ = 0, then it is not correct. é log (1 + x ) ù y®0 èy yø êëQ xlim = 1ú ®0 x û 258 Textbook of Differential Calculus Since, limit is finite, so (a – b ) = 0 Þ b = a Exponential Limits ax 2 There are two types of exponential limits discussed below xa + +... ¥ 2! \ lim =2 x ®0 x (i) Based on Series Expansion ax Þ lim a + +... ¥ = 2 Þ a = 2 x2 x 3 x ®0 2! ex =1+ x + + +... ¥ 2! 3! \ b=2 To evaluate the exponential limit, we use the following a sin x - bx + cx 2 + x 3 results. y Example 38 Solve lim , if it log (1 + x ) - 2x 3 + x 4 x ®0 2x 2 ex –1 ax –1 (a) lim =1 (b) lim = log e a exists and is finite, also find a, b and c. x ®0 x x ®0 x a sin x - bx + cx 2 + x 3 Sol. Let L = lim ax – b x x ® 0 2x 2 log (1 + x ) - 2x 3 + x 4 y Example 34 Evaluate lim × x ®0 x æ x3 x5 ö aç x - + -...÷ - bx + cx 2 + x 3 ax – bx ( a x –1 ) – ( b x –1 ) è 3 ! 5! ø Sol. We have, lim = lim = lim x ®0 x x ®0 x x ®0 æ x2 x3 x4 ö x x 2x 2 ç x - + - +...÷ - 2x 3 + x 4 a –1 b –1 è 2 3 4 ø = lim – lim = log a – log b = log (a / b ) x ®0 x x ®0 x æ aö a (a - b )x + cx 2 + ç1 - ÷ x 3 + x 5 +... (ab ) x – a x – b x + 1 è 3! ø 5! y Example 35 Evaluate lim × = lim x ®0 2 5 1 6 x ®0 x2 x - x +... 3 2 (ab )x – a x –b x + 1 a x b x – a x –b x + 1 Sol. We have, lim = lim a x ®0 x 2 x ® 0 x2 For finite limit, a - b = 0, c = 0, 1 - = 0, i.e. a = b = 6, c = 0] x x a (b – 1)–(b –1) x 3! = lim a x ®0 x2 + higher powers of x x ( a – 1) ( b x –1 ) \ L = lim ! 5 = lim ´ lim = log a ´ log b x ®0 2 1 - x +L x ®0 x x ®0 x 3 2 é ax - 1 ù a 3 a 6 3 ê x ®0 Q lim = loge a ú = × = = = [Qa = 6] ë x û 5! 2 80 80 40 a sin x - bx + cx 2 + x 3 3 e tan x – e x So, lim = y Example 36 Evaluate lim × x ®0 2x log (1 + x ) - 2x + x 2 3 4 40 x ® 0 tan x – x Where, a = 6 = b, c = 0 e tan x – e x e x ´ e (tan x – x ) – e x Sol. We have, lim = lim y Example 39 Find the values of a, b and c such that x ® 0 tan x – x x ®0 ( tan x – x ) e x {e tan x – x – 1} axe x – b log (1 + x ) + cxe – x = lim x ®0 (tan x – x ) lim = 2. x ® 0 x 2 sin x = e 0 ´ 1 [as x ® 0, tan x – x ® 0] axe x - b log (1 + x ) + c xe - x =1 ´ 1 =1 Sol. We have, lim =2 x ®0 x 2 sin x ae x – b Using the expansion, we have y Example 37 Evaluate lim = 2. Find a and b. x ®0 x é æ x2 ö æ x2 x3 ö ù êax ç1 + x + 2 ! +...÷ –b ç x – 2 + 3 –...÷ ú ê è æ x x2 ö ø è ø ú a ç1 + + +... ¥ ÷ –b è ø ê æ x x2 x3 öú ae x - b 1! 2! ê + cx ç1 – + - +...÷ ú Sol. Given, lim = lim =2 x ®0 x x ®0 x êë è 1! 2! 3! øú lim û =2 ax 2 x ®0 æ 3 5 ö (a –b ) + xa + +... ¥ x 2 çx – x + x –...÷ 2! è ø Þ lim =2 3! 5! x ®0 x  Chap 05 Limits 259 æ ö px b æa b c ö tan y Example 42 Evaluate lim æç 2 – ö÷ x (a – b + c )– x 2 çça + – c ÷÷ + x 3 ç – + ÷ +... a 2a è 2 ø è2 3 2ø. x®a è xø lim =2 x ®0 æ x x 3 ö 5 px x 2 çx – + –...÷ tan è 3! 5! ø æ aö 2a Sol. We have, lim ç2 – ÷ x ®a è xø Now, above limit would exist, if least power in numerator is