13.1 Overview 13.1.1 Limits of a function Let f be a function defined in a domain which we take to be an interval, say, I. We shall study the concept of limit of f at a point ‘a’ in I. We say lim fx( ) is the expected value of f at x = a given the values of f near to the xa –→left of a. This value is called the left hand limit of f at a. fxWe say lim ( ) is the expected value of f at x = a given the values of f near to the→+xa right of a. This value is called the right hand limit of f at a. If the right and left hand limits coincide, we call the common value as the limit of f at x = a and denote it by lim ( ) .fx xa→ Some properties of limits Let f and g be two functions such that both lim ( ) and lim gxfx ( ) exist. Then xa→xa→()] () gx (i) lim[() fx + gx = lim fx + lim () xa xa →a→ → x ()] () gx (ii) lim[ ( ) fx − gx = lim fx − lim () xa xa →a→ → x (iii) For every real number α lim ( α )( ) =α lim f () fx x xa x→→ a (iv) lim [ fxg () x = [lim fx gx ()] ()lim ()] xa→ xa→ xa→ lim () fx() xafx →lim = , provided g (x) ≠ 0gx ()xa→ () lim gx xa→ Limits of polynomials and rational functions If fis a polynomial function, then lim fx → ( ) exists and is given byxa lim fx() = fa () xa→ An Important limit An important limit which is very useful and used in the sequel is given below: xn− an n− 1lim = na xa xa→− Remark The above expression remains valid for any rational number provided ‘a’ is positive. Limits of trigonometric functions To evaluate the limits of trigonometric functions, we shall make use of the following limits which are given below: sin x(i) lim = 1 (ii) lim cos x=1 (iii) lim sin x= 0 x→0 x→0x→0 x 13.1.2 Derivatives Suppose fis a real valued function, then ( + ) − fxfx h () f′(x) = lim ... (1)h→0 h is called the derivative of fat x, provided the limit on the R.H.S. of (1) exists. Algebra of derivative of functions Since the very definition of derivatives involve limits in a rather direct fashion, we expect the rules of derivatives to follow closely that of limits as given below: Let fand gbe two functions such that their derivatives are defined in a common domain. Then : (i) Derivative of the sum of two function is the sum of the derivatives of the functions. d dd[ () +gx () ] fx+ ()fx = () gx dx dxdx (ii) Derivative of the difference of two functions is the difference of the derivatives of the functions. d dd[ () −() ] = fx− ()fx gx () gx dx dxdx LIMITS AND DERIVATIVES 227 (iii) Derivative of the product of two functions is given by the following product rule. d ⎛d ⎞⎛d ⎞fx () gx ⎟[ () ⋅gx () ]= ⎜ fx()⎟⋅gx () +fx ⋅⎜ ()dx ⎝dx ⎠⎝dx ⎠ This is referred to as Leibnitz Rule for the product of two functions. (iv) Derivative of quotient of two functions is given by the following quotient rule (wherever the denominator is non-zero). ⎛d ⎞⎛d ⎞() ⋅() −fx ⋅ ()⎜ fx ⎟gx () ⎜ gx ⎟d ⎛f ()x ⎞⎝dx ⎠⎝dx ⎠ ⎜⎟ 2 = dx ⎝ () ()gx ⎠gx()13.2 Solved Examples Short Answer Type ⎡1 2(2 x −3) ⎤Example 1 Evaluate lim −⎢ 32 ⎥x→2 x −2 x −3x +2x⎣ ⎦ Solution We have ⎡1 2(2 x −3) ⎤⎡1 2(2 x −3) ⎤lim − lim − →2 ⎢ 32 ⎥= ⎢x ⎣x −2 x −3x +2x⎦x→⎣ 2 xx −1) ( x −2⎥⎦2 x − ( xx( 1) 2(2 x −3) ⎤⎡ −−lim= ⎢ ⎥→2 ⎣ ( −1) ( x −2) ⎦x xx ⎡x2 −+ 5x 6 ⎤ = lim ⎢⎥ x→2 ⎣xx( −1) ( x −2) ⎦ ⎡( x −2) ( x −3) ⎤lim= ⎢⎥[x – 2 ≠ 0]→2 ⎣ ( −1) ( x −2) ⎦x xx ⎡x −3 ⎤−1lim = = x→2 ⎢⎣xx( −1) ⎥⎦ 2 Example 2 Evaluate lim x→0 x Solution Put y = 2+ x so that when x → 0, y → 2. Then 1 1 0x → x = 2 lim y→ 2y y 22 2 − − = 1 1 21 (2) 2 − 1 2 2 − =⋅ 1 2 = 1 2 2 nx 3n− Example 3 Find the positive integer n so that lim 108 = . x →3 x −3 Solution We have xn −3n lim n(3)n – 1= x →3 x −3 n(3)n – 1Therefore, = 108 = 4 (27) = 4(3)4 – 1 Comparing, we get n =4 Example 4 Evaluate lim (sec x −tan x)π x →2 ππSolution Put y = −x . Then y → 0 as x → . Therefore22 ππlim (sec x −tan x) = lim [sec( −− y) tan ( −y)] π y→0x→2 22 lim (cosec y −cot y)= y→0⎛1 cos y⎞ = lim −⎜⎟y→0 ⎝sin y sin y ⎠ 1cos y⎞⎛− = lim ⎜⎟y→0 ⎝sin y ⎠ LIMITS AND DERIVATIVES 229 2 y ⎛ 2 y 1cos − y⎞2sin since , sin = 2 ⎜ 22 ⎟ = lim ⎜⎟ y→0 yy yy2sin cos ⎜sin y=2sin cos ⎟22 ⎝ 22 ⎠ y= lim tan = 0 y 2→02 sin (2 x) sin(2 −x)+− Example 5 Evaluate lim x→0 x Solution (i) We have (2 x 2 x) (2 x 2 x)++− +−+ 2cos sin sin (2 x) sin(2 −x) 22+− lim = lim x→0 x→0xx 2cos 2 sin x = lim x→0 x sin x ⎛ sin x ⎞= 2 cos2 lim =2cos 2 ⎜as lim =1⎟ x→0 x ⎝ x→0 x ⎠ Example 6 Find the derivative of f(x) = ax+ b, where aand bare non-zero constants, by first principle. Solution By definition, f′(x) = 0 lim h→ = 0 lim h→ fx h () fx+− () h () b (a+) bhax h + +− xb = lim = bhh→0 hExample 7 Find the derivative of f(x) = ax2 + bx+ c, where a, band care none-zero constant, by first principle. Solution By definition, fx h () fx+− () f′(x) = lim h→0 h 230 EXEMPLAR PROBLEMS – MATHEMATICS lim axh 2( + ca2 xc( + ) + bxh ) +− x − b− = h→ 0 h bh ah +2 h = lim + 2 ax = lim ah+ 2ax+ b= b+ 2axh→0 hh→0 Example 8 Find the derivative of f(x) = x3, by first principle. Solution By definition, ( + ) − fxfx h () f′(x) = lim h→0 Example 9 Find the derivative of fSolution By definition, h (x+ h)3 − x3 = lim h→0 h 33 3+ h+ 3( + )x xhx h − x = lim h→0 h lim 2= h→0(h2 + 3x(x+ h)) = 3x1 (x) = by first principle.x( + ) − fxfx h () f′(x) = lim h→0 h 1 ⎛ 11⎞ = lim −⎜⎟h→0 hxh + x⎝⎠ −h −1 =lim = .h→0( +)2hxhx x Example 10 Find the derivative of f(x) = sin x, by first principle. Solution By definition, ( + ) − fxfx h () f′(x) = lim h→0 h LIMITS AND DERIVATIVES 231 sin ( x + h) − sin xlim= h→0 h ⎛ 2x + h ⎞ h2cos ⎜⎟ sin ⎝⎠22 = lim h→0 h2 ⋅ 2 hsin (2 + )2xh = lim cos ⋅ lim h→02 h→0 h 2 = cos x.1 = cos x Example 11 Find the derivative of f (x) = xn , where n is positive integer, by first principle. Solution By definition, − () fx( + h) fxf ′(x) = h (x + h)n − xn = h Using Binomial theorem, we have (x + h)n = nCxn + nCxn – 1 h + ... + n C hn 01 n (x + h)n − xn limThus, f ′(x) = h→0 h n−1 n −1( + ... +hhnx ]lim = nxn – 1= .h→0 h Example 12 Find the derivative of 2x4 + x. Solution Let y = 2x4 + x Differentiating both sides with respect to x, we get dy d 4 d(2 x ) + () x= dxdx dx =2 × 4x4 – 1 + 1x0 =8x3 + 1 d Therefore, dx (2 x4 +x) =8x3 + 1. Example 13 Find the derivative of x2 cosx. Solution Let y = x2 cosx Differentiating both sides with respect to x, we get dy d 2 = (cos x)x dxdx 2 dd 2x (cos x) + cos x (x )= dx dx = x2 (– sinx) + cosx (2x) =2x cosx – x2 sinx Long Answer Type 2sin 2 x + sin x − 1Example 14 Evaluate lim 2 x →π 2sin x − 3sin x + 1 6 Solution Note that 2 sin2 x + sin x – 1 = (2 sin x – 1) (sin x + 1) 2 sin2 x – 3 sin x + 1 = (2 sin x – 1) (sin x – 1) 22sin x + sin x − 1 (2sin x −1) (sin x +1) Therefore, lim 2 = lim x π 2sin x − 3sin x +1 π (2sin x −1) (sin x −1) → x →66 sin x + 1 = lim (as 2 sin x – 1 ≠ 0) x →π sin x − 1 6 π1sin + 6 = = –3πsin −16 tan x−sin xExample 15 Evaluate lim x→0 sin 3 x Solution We have tan x−sin xlim 3= x→0 sin x = = LIMITS AND DERIVATIVES 233 ⎛1 ⎞sin x⎜−1⎟⎝⎠cos xlim 3x→0 sin x 1cos x−lim x→0 cos xsin 2 x x2sin 2 12lim = x→0 ⎛ 2 x 2 x⎞ 2. cos x⎜4 sin ⋅cos ⎟⎝ 22⎠ax 23+− x Example 16 Evaluate lim xa 3ax2→ +− x ax 23+− xSolution We have lim xa→ 3ax2+− x = = (ax) (−lim= +− 23xa2 x 3ax ++ xlim × xa 3ax2 a2 x 3x→ +− x ++ 23axx +− lim xa ax→(3 +−2 x)(ax 3x++ 2 ) ++ 2 x)3ax xa→(a++ 2 x 3x)(3ax +− 2 x)(3ax ++ 2 x) (ax−) ⎡3ax 2++ x⎤⎣ ⎦lim= →(+ + 2 x 3 )(3 +− 4 )xaa xax x 4 a 22 3 = = = .323a 33 9×cos ax−cos bxExample 17 Evaluate lim x→0 cos cx−1 ⎛(ab+)( −)⎞ abx2sin ⎜ x⎟sin ⎝ 2 ⎠ 2Solution We have lim x→0 sin 2 cx2 2 (ab) x ( −)+ abx2sin ⋅sin 2x22 = lim 2 ⋅ x→0 cxx 2sin 2 (abx (ab) x ⎛⎞24+) − cxsin sin ⎜⎟×⎝⎠ 22 22 c = lim ⋅⋅ x→0( +) x⎛2(ab−) x 22 cxab⎞⋅ ⋅ sin ⎜⎟⎝+⎠ 2 ab− 22 ab ab ab − 4 ⎞ 2 −2⎛+ ab = ⎜ 2 ⎟=⎝2 × 2 × c⎠c2 ( +)2 sin( ah a ) 2sin ah +− aExample 18 Evaluate lim h→0 h ( +)2 sin( ah a ) 2sin ah +− aSolution We have lim h→0 h 22 2(ah 2a)[sin acos h+cos asin ha sin ++ h ] −a = lim h→0 h a2 sin a(cos h−1) a2 cos asin h = lim [ + (ha acos h+cos asin ++ 2 ) (sin h)] h→0 h h LIMITS AND DERIVATIVES 235 ⎡ h ⎤2 sin a ( 2sin 2 ) 2a −⎢ 2 h⎥ a cos a sin h 2 a + h)lim ⎢ ⋅⎥+ lim + lim ( h + 2 )sin( a h→0 h 2 h→0 hh→0= ⎢⎥ ⎣ 2 ⎦⎢ ⎥ = a2 sin a × 0 + a2 cos a (1) + 2a sin a = a2 cos a + 2a sin a. Example 19 Find the derivative of f (x) = tan (ax + b), by first principle. − () fx( + h) fx Solution We have f ′(x) = lim h→0 h lim tan (ax( + h) + b)− tan ( ax + b) = h→0 h sin ( ax + ah + b) sin ( ax + b)−cos ( ax + ah + b) cos( ax + b)lim= h→0 h sin ( ax + ah + b) cos ( ax + b) − sin ( ax + b) cos ( ax + ah + b) = lim h→0 h cos ( ax + b) cos ( ax + ah + b) a sin ( ah ) = lim h→0 ah⋅ cos ( ax + b) cos ( ax + ah + b) a sin ah = lim lim [as h → 0 ah → 0]h→ ah 0 cos ( ax + b) cos ( ax + ah +b) → 0 ah a = 2 = a sec2 (ax + b).cos (ax + b) fxExample 20 Find the derivative of () = sin x , by first principle. Solution By definition, − () fx( + h) fxf ′(x) = lim h→0 h = h→0 h ( sin ( x + h) − sin x )( sin ( x + h) + sin x)= lim h→0 h ( sin ( x + h) + sin x) sin ( xh+ ) − sin xlim= h→0 h( ⎛ 2xh+⎞ h2cos ⎜⎟ sin ⎝⎠22 = lim cos x 1 = = cot x sin x 2sin x2 cos xExample 21 Find the derivative of .1sin x+ cos x Solution Let y = 1sin x+ Differentiating both sides with respects to x, we get dy d ⎛ cos x ⎞ = ⎜⎟dx dx ⎝ 1sin + x⎠dd(1 + sin x) (cos x) − cos x (1 + sin x)dx dx = (1+ sin x)2 (1 + sin x)( −sin x) − cos x (cos x)= (1 + sin x)2 LIMITS AND DERIVATIVES 237 −sin x −sin 2 x −cos 2 x = (1 +sin x)2 (1 sin x −−+ )1 = = (1 +sin x)2 1sin x+ Objective Type Questions Choose the correct answer out of the four options given against each Example 22 to 28 (M.C.Q.). sin xExample22 lim is equal tox→0 x(1 +cos x) 1 (A) 0 (B) (C)1 (D) –1 2 Solution (B) is the correct answer, we have xx2sin cos sin x 22lim lim= x →0 x(1 +cos x) x →0 ⎛ x⎞ x ⎜2cos 2 ⎟⎝ 2⎠ xtan 112= lim = 2 x →0 x 2 2 1sin x−Example23 lim is equal toπ cos xx →2 (A) 0 (B) –1 (C) 1 (D) does not exit Solution (A) is the correct answer, since ⎡ ⎛π ⎞⎤ ⎢1sin y⎟⎥1sin x 2 ⎠⎛ πlim − = lim − ⎝⎜−taking −=y ⎞ x →π cos xy→0 ⎢⎢ ⎛π ⎞⎥⎥⎝⎜ 2 x ⎠⎟ 2 cos ⎜−y⎟⎢⎣ ⎝2 ⎠⎥⎦ 238 EXEMPLAR PROBLEMS – MATHEMATICS 2 y2sin lim 1cos −y lim 2 = = y→0 yyy→0 sin y 2sin cos 22 y= lim tan = 0 y→02 ||xExample 24 lim is equal tox →0 x(A) 1 (B) –1 (C) 0 (D) does not exists Solution (D) is the correct answer, since || xx R.H.S = lim + ==1 x→0 xx || −xxand L.H.S = lim = =−1 x →0– xx Example 25 lim [ x −1] , where [.] is greatest integer function, is equal tox→1(A) 1 (B) 2 (C) 0 (D) does not exists Solution (D) is the correct answer, since R.H.S = lim [ x −=01] x →1+ lim [ x 1] −=–1 and L.H.S = −x →11Example 26 lim x sin is equals tox →0 x 1 (A) 0 (B) 1 (C) (D) does not exist 2 Solution (A) is the correct answer, since 1lim x =0 and –1 ≤ sin ≤ 1, by Sandwitch Theorem, we havex →0 x LIMITS AND DERIVATIVES 239 1lim xsin =0 x→0 x 1 +++ + 2 3 ... n Example 27 lim 2, n∈ N, is equal ton→∞ n 11(A) 0 (B)1 (C) (D)24 1 +++ + 2 3 ... n Solution (C) is the correct answer. As lim 2x→∞ n ( +1) 1 ⎞ 1nn 1 ⎛lim lim ⎜1+ ⎟= = 2 = ⎝⎠n→∞ 2nx→∞2 n 2 ⎛⎞fπ′Example 28 If f(x) = xsinx, then ⎜⎟is equal to⎝⎠2 1(A) 0 (B)1 (C)–1 (D) 2 Solution (B) is the correct answer. As f′ (x) = xcosx+ sinx π⎛⎞ππ π So, f′⎜⎟= cos +sin =1⎝⎠222 2 Short Answer Type Evaluate : 1. 2 3 9lim 3x x → x − − 2. 2 1 2 4 1lim 2 1x x x→ − − 3. 0 lim h→ xh h +− x 4. 1 3 0 ( 2) lim x x x→ + 1 32− 5. 1 (1lim (1x→ + + 6 2 ) 1 ) 1 x x − − 6. 5 5 2 2(2 ) ( 2) lim xa x a xa→ + − + − 4 2 x − xx − 4lim lim7. 8. x→1 x − 1 x →23x −−2 x + 2 4 75 33x − 4 x − 2x + 11 + x − 1 − x9. lim 2 10. lim3 2 11. lim 2x → 2 x + 32 x − 8 x→1 x − 3x + 2 x →0 x x3 + 27 ⎛ 8x − 34x2 + 1⎞ 12. lim 13. lim − x →−3 1 ⎜ 2 ⎟x5 + 243 x →⎝ 2x −14x −1⎠ 2 14. Find ‘n’, if 2 lim x → nx x − − 2 2 n 80 = , n ∈ N 15. sin 3 lim xa sin 7 x x→ 16. 0 lim x→ 2 2 sin 2 sin 4 x x 17. 20 1cos2 lim x x x→ − 18. 0 lim x → 3 2sin sin 2 x x x − 19. 0 1cos lim 1cos x mx nx→ − − 20. 3 lim x π→ 1cos 6 2 3 x x − ⎛ π ⎞−⎜ ⎟⎝ ⎠ 21. 4 lim x π→ sin x x − − cos 4 x π 22. lim π 3sin x cos − π x 23. 0 lim x → sin 2 3 2 tan3 x x x x + + 24. lim xa→ sin x x sin a a − − x → x −66 cot 2 x − 325. lim 26. lim 2π x →0 x → cosec x − 2 sin x 6 sin x − 2sin3 x + sin5 x27. lim x→0 x 4 33x −1 x − k28. If lim = lim 2 2 , then find the value of k. x→1 x − 1 x →kx − k Differentiate each of the functions w. r. to x in Exercises 29 to 42. 4 32 3 x + x + x + 1 ⎛ 1 ⎞29. 30. ⎜ x +⎟ 31. (3x + 5) (1 + tanx)⎝ ⎠x x LIMITS AND DERIVATIVES 241 3x +4 x5 −cos x32. (sec x – 1) (sec x + 1) 33. 2 34.5x −7 x +9 sin x 2 π x cos 435. 36. (ax2 + cotx) (p + q cosx)sin x ab+ sin x 37. 38. (sin x + cosx)2 39. (2x – 7)2 (3x + 5)3 cd+ cos x 1 40. x2 sinx + cos2x 41. sin3x cos3x 42. 2ax +bx +c Long Answer Type Differentiate each of the functions with respect to ‘x’ in Exercises 43 to 46 using first principle. ax +b 243. cos (x2 + 1) 44. 45. x3cx +d 46. x cosx Evaluate each of the following limits in Exercises 47 to 53. ( x +y) sec( x +y) −xsec x 47. lim y→0 y (sin( α+β ) x +sin( α−β ) x +sin 2 αx)lim ⋅x48. x →0 cos 2 β−x cos 2 αx x1sin − x 2tan 3 x −tan lim49. lim 50. x →π x ⎛ xx⎞ x →π ⎛π⎞ cos ⎜cos −sin ⎟4 cos ⎜x +⎟ 2 ⎝ 44⎠⎝ 4 ⎠ | x −4| 51. Show that lim does not exists x →4 x −4 ⎧kcos x πwhen x≠⎪π−2 x 2 π52. Let f(x) = ⎨ and if lim ( ) = f( ), ⎪ π x→π 23 x= 2 fx ⎪ 2⎩find the value of k. ⎧x+2 x≤–1 fx53. Let f(x) = , find ‘c’ if lim ( ) exists.⎨ 2 x→–1 cx x>−1⎩Objective Type Questions Choose the correct answer out of 4 options given against each Exercise 54 to 76 (M.C.Q). sin x 54. lim is x→π x−π (A) 1 (B)2 (C)–1 (D) –2 2xcos x55. lim is x→0 1cos x−3 −3 (A) 2 (B) (C) (D)122 (1+ x)n−156. lim is x→0 x (A) n (B)1 (C)–n (D) 0 xm−157. lim n is x→1 x−1 2m mm(A) 1 (B) (C) − (D) 2n nn 1cos4θ−58. lim is →0 1cos6θx− LIMITS AND DERIVATIVES 243 41 −1(A) (B) (C) (D) –1 92 2 cosec x−cot x 59. lim is x→0 x −1 1 (A) (B)1 (C) (D)1 22 sin xlim60. − is x→0 x 1+−1 x (A) 2 (B)0 (C)1 (D) –1 sec 2 x−2 61. lim is x→πtan x−1 4 (A) 3 (B)1 (C)0 (D) 2 (x−12(x)−3)62. lim2 is x→12x+−x 3 1 −1 (A) (B) (C) 1 (D) None of these 10 10 ⎧sin[ ] x,[ ] ≠0x⎪⎨[] ⎪ 63. If f(x) = x , where [.] denotes the greatest integer function , ⎩0 ,[] x=0 lim () then x→0 fx is equal to (A) 1 (B) 0 (C) –1 (D) None of these |sin x|64. lim is x→0 x (A) 1 (B) –1 (C) does not exist(D) None of these ⎧x2 −1, 0 << x 2 fx65. Let f(x) =⎨ ,the quadratic equation whose roots are lim ( ) and ⎩ x 3 2–2x+3, 2 ≤< x→lim () fx x→2+ is (A) x2 – 6x + 9 = 0 (B) x2 – 7x + 8 = 0 (C) x2 – 14x + 49 = 0 (D) x2 – 10x + 21 = 0 tan 2 x −x66. lim is x →03x −sin x 1 −11 (A) 2 (B) (C) (D)2 24 1⎛⎞67. Let f (x) = x – [x]; ∈ R, then f ′⎝⎠is⎜⎟2 3 (A) (B)1 (C)0 (D) –1 2 1 dy68. If y = x + , then at x = 1 is x dx1 1 (A) 1 (B) (C) (D) 022 x −4 69. If f (x) = , then f ′(1) is2 x 54(A) (B) (C)1 (D)0 45 11 +2 dy70. If y = x, then dx is11 −2x −4x −4 x 1 −x24x (A) 2 2(B)2 (C) (D)2( x −1) x −14xx −1 sin x +cos x dy71. If y = , then at x = 0 issin x −cos x dxLIMITS AND DERIVATIVES 245 1 (A) –2 (B) 0 (C) (D) does not exist 2 sin( x +9) dy72. If y = then at x = 0 is cos x dx(A) cos 9 (B) sin 9 (C) 0 (D) 1 2 100 xx ++ +73. If f (x) = 1 x+... , then f ′(1) is equal to2 100 1 (A) (B) 100 (C) does not exist (D) 0 100xn −an fx74. If () = for some constant ‘a’, then f ′(a) isx −a 1 (A) 1 (B) 0 (C) does not exist (D) 2 75. If f (x) = x100 + x99 + ... + x + 1, then f ′(1) is equal to (A) 5050 (B) 5049 (C) 5051 (D) 50051 76. If f (x) = 1 – x + x2 – x3 ... – x99 + x100, then f ′(1) is euqal to (A) 150 (B) –50 (C) –150 (D) 50 Fill in the blanks in Exercises 77 to 80. tan x77. If , then lim ( ) f (x) = x −πx→πfx = ______________ ⎛ x ⎞78. lim sin mx cot =2 , then m = ______________⎜ ⎟ x →0 ⎝ ⎠ 23 dyxx x79. if y =1 ++ + +..., then = ______________1!2! 3! dx80. lim = ______________ x →3+[]x

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