Chapter 13 – Limits And Derivatives

Class 11 Mathematics · 54 questions · 0 with answers

Solved examples

example-1Long answer

Evaluate lim  x→ 2 x − 2 − 3 x − 3 x 2 + 2 x   Solution We have  1 2 (2 x − 3)   1 2 (2 x − 3)  lim  − 3 2  = lim  − x→ 2 x − 2  x − 3x + 2 x  x→ 2 x − 2  x ( x − 1) ( x − 2   x ( x − 1) − 2 (2 x − 3)  = lim x→ 2    x ( x − 1) ( x − 2)   x2 − 5x + 6  = x→ 2    x ( x − 1) ( x − 2)   ( x − 2) ( x − 3)  = lim x → 2  x ( x − 1) ( x − 2)  [x – 2 ≠ 0]    x − 3  −1 = lim x → 2  x ( x − 1)  =   2 2+ x − 2

example-2Short answer

Evaluate lim x→0 x Solution Put y = 2 + x so that when x → 0, y → 2. Then 1 1 2+ x − 2 y − 22 lim = lim x→0 x y→2 y − 2 1 1 1 2 −1 1 − 2 1 = (2) = ⋅2 = 2 2 2 2 x n − 3n

example-3Short answer

Find the positive integer n so that lim = 108 . x →3 x − 3 Solution We have x n − 3n lim = n(3)n – 1 x →3 x − 3 Therefore, n(3)n – 1 = 108 = 4 (27) = 4(3)4 – 1 Comparing, we get n= 4

example-4Short answer

Evaluate lim (sec x − tan x) π x→ π π Solution Put y = − x . Then y → 0 as x → . Therefore

example-5Short answer

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) 2 2 lim = lim x→0 x x→0 x 2cos 2 sin x = lim x→0 x sin x  sin x  = 2cos 2  as lim = 2 cos 2 lim = 1  x→0 x x→0 x

example-6Short answer

Find the derivative of f (x) = ax + b, where a and b are non-zero constants, by first principle. Solution By definition, f ( x + h) − f ( x ) f ′(x) = lim h→0 h a ( x + h) + b − (ax + b) bh = lim = lim =b h→0 h h →0 h

example-7Short answer

Find the derivative of f (x) = ax2 + bx + c, where a, b and c are none-zero constant, by first principle. Solution By definition, f ( x + h) − f ( x ) f ′(x) = lim h→0 h a ( x + h) 2 + b ( x + h) + c − ax 2 − bx − c = lim h→ 0 h bh + ah 2 + 2axh = lim = lim h→0 ah + 2ax + b = b + 2ax h→ 0 h

example-8Short answer

Find the derivative of f (x) = x3, by first principle. Solution By definition, f ( x + h) − f ( x ) f ′(x) = lim h→0 h ( x + h )3 − x 3 = lim h→ 0 h x 3 + h3 + 3 xh ( x + h) − x 3 = lim h→ 0 h = lim h→0 (h2 + 3x (x + h)) = 3x2

example-9Short answer

Find the derivative of f (x) = by first principle. Solution By definition, f ( x + h) − f ( x ) f ′(x) = lim h→0 h 1 1 1 = lim h→0 h  −   x + h x −h −1 = lim = 2. h→0 h ( x + h) x x

example-10Short answer

Find the derivative of f (x) = sin x, by first principle. Solution By definition, f ( x + h) − f ( x ) f ′(x) = lim h→0 h LIMITS AND DERIVATIVES 231 sin ( x + h) − sin x = lim h→0 h  2x + h  h 2 cos  sin  2  2 = h→0 h 2⋅ (2 x + h) 2 = lim cos ⋅ lim h→ 0 2 h→0 h = cos x.1 = cos x

example-11Short answer

Find the derivative of f (x) = xn, where n is positive integer, by first principle. Solution By definition, f ( x + h) − f ( x ) f ′(x) = ( x + h) n − x n = Using Binomial theorem, we have (x + h)n = nC0 xn + nC1 xn – 1 h + ... + nCn hn ( x + h) n − x n Thus, f ′(x) = lim h→ 0 h h (nx n −1 + ... + h n −1 ] = lim = nxn – 1. h→ 0 h

example-12Short answer

Find the derivative of 2x4 + x. Solution Let y = 2x4 + x Differentiating both sides with respect to x, we get dy d d = (2 x 4 ) + ( x) dx dx dx = 2 × 4x4 – 1 + 1x0 = 8x3 + 1 Therefore, (2 x 4 + x) = 8x3 + 1.

example-13Short answer

Find the derivative of x2 cosx. Solution Let y = x2 cosx Differentiating both sides with respect to x, we get dy d 2 = ( x cos x) dx dx d d 2 = x2 (cos x) + cos x (x ) dx dx = x2 (– sinx) + cosx (2x) = 2x cosx – x2 sinx

example-14Short answer

Evaluate limπ x→ 2sin 2 x − 3sin x + 1 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) 2sin 2 x + sin x − 1 (2sin x − 1) (sin x + 1) Therefore, lim 2 = lim x → 2sin x − 3sin x + 1 x → (2sin x − 1) (sin x − 1) π π

example-15Short answer

Evaluate lim x→0 sin 3 x Solution We have  1  tan x − sin x sin x  − 1 lim = lim  cos x  x→0 sin 3 x x→0 sin 3 x 1 − cos x = lim x →0 cos x sin 2 x 2 sin 2 lim 2 1 = x→0  x x = . cos x  4 sin 2 ⋅ cos 2  2  2 2 a + 2 x − 3x

example-16Long answer

Evaluate lim x→ a 3a + x − 2 x a + 2 x − 3x Solution We have lim x→ a 3a + x − 2 x a + 2 x − 3x a + 2 x + 3x = lim × x→ a 3a + x − 2 x a + 2 x + 3x a + 2 x − 3x = lim x→a ( 3a + x − 2 x )( a + 2 x + 3x ) = lim ( 3a + x + 2 x ) (a − x) x→ a ( a + 2 x + 3x )( 3a + x − 2 x )( 3a + x + 2 x ) (a − x) 3a + x + 2 x = lim x→ a ( a + 2 x + 3x ) (3a + x − 4 x) 4 a 2 2 3 = = = . 3 × 2 3a 3 3 9 cos ax − cos bx

example-17Short answer

Evaluate lim x→0 cos cx − 1  ( a + b)  ( a − b) x 2 sin  x sin  2  2 Solution We have x→0 sin 2 cx ( a + b) x ( a − b) x 2sin ⋅ sin 2 2 x2 = lim ⋅ x→0 x2 sin 2 ( a + b) x ( a − b) x  cx  4 sin sin   × 2 lim 2 ⋅ 2 ⋅ 2 c = x → 0 ( a + b ) x  2  ( a − b) x 2 cx ⋅ ⋅ sin 2 2  a + b  2 a−b 2 a+b a−b 4  a 2 − b2 =  × × 2 = 2 2 c  c2 ( a + h) 2 sin ( a + h) − a 2 sin a

example-18Long answer

Evaluate lim h→ 0 h ( a + h) 2 sin ( a + h) − a 2 sin a Solution We have lim h→ 0 h ( a 2 + h 2 + 2ah) [sin a cos h + cos a sin h] − a 2 sin a = lim h→ 0 h a 2 sin a (cos h − 1) a 2 cos a sin h = lim [ + + (h + 2a ) (sin a cos h + cos a sin h)] h→ 0 h h LIMITS AND DERIVATIVES 235  2 2 h   a sin a ( −2 sin 2 ) h  a 2 cos a sin h lim  ⋅  + lim + lim (h + 2a ) sin (a + h) = h→0  h2 2  h→0 h h→ 0  2  = a2 sin a × 0 + a2 cos a (1) + 2a sin a = a2 cos a + 2a sin a.

example-19Long answer

Find the derivative of f (x) = tan (ax + b), by first principle. f ( x + h) − f ( x ) Solution We have f ′(x) = lim h→0 h tan ( a ( x + h) + b ) − tan (ax + b) = lim 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 a ⋅ h cos (ax + b) cos (ax + ah + b) a sin ah = lim lim [as h → 0 ah → 0] h→0 cos ( ax + b) cos ( ax + ah + b) ah→ 0 ah = 2 = a sec2 (ax + b). cos (ax + b)

example-20Long answer

Find the derivative of f ( x) = sin x , by first principle. Solution By definition, f ( x + h) − f ( x ) f ′(x) = lim h→0 h sin ( x + h) − sin x = lim h→ 0 h ( sin ( x + h) − sin x )( sin ( x + h) + sin x ) = lim h→ 0 h ( sin ( x + h) + sin x ) sin ( x + h) − sin x = lim h→ 0 h ( sin ( x + h) + sin x )  2x + h  h 2 cos  sin  2  2 = h→ 0 h 2⋅ ( sin ( x + h) + sin x ) cos x 1 = = cot x sin x 2 sin x 2 cos x

example-21Short answer

Find the derivative of . 1 + sin x cos x Solution Let y = 1 + sin x Differentiating both sides with respects to x, we get dy d  cos x  = dx  1 + sin x  dx   d d (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 = 2 = (1 + sin x) 1 + sin x

example-24Multiple choice

lim is equal to x→0 x

  • (A)1
  • (B)–1
  • (C)0
  • (D)does not exists Solution (D) is the correct answer, since | x| x R.H.S = lim+ = =1 x →0 x x | x | −x and L.H.S = lim– = = −1 x→0 x x
example-25Multiple choice

lim x →1 [ x − 1] , where [.] is greatest integer function, is equal to

  • (A)1
  • (B)2
  • (C)0
  • (D)does not exists Solution (D) is the correct answer, since R.H.S = lim [ x − 1] = 0 x →1+ and L.H.S = lim [ x − 1] = –1 x →1−
example-26Multiple choice

lim x sin is equals to x →0 x

  • (A)0
  • (B)1
  • (C)
  • (D)does not exist Solution (A) is the correct answer, since lim x = 0 and –1 ≤ sin x→0 ≤ 1, by Sandwitch Theorem, we have LIMITS AND DERIVATIVES 239 lim x sin = 0 x →0 x 1 + 2 + 3 + ... + n
example-27Multiple choice

nlim , n ∈ N, is equal to →∞ n2 1 1

  • (A)0
  • (B)1
  • (C)
  • (D)2 4 1 + 2 + 3 + ... + n Solution (C) is the correct answer. As xlim →∞ n2 n (n + 1) 1 1 1 = nlim = lim 1 +  = →∞ 2n 2 x →∞ 2 n 2  π
example-28Multiple choice

If f(x) = x sinx, then f ′   is equal to

  • (A)0
  • (B)1
  • (C)–1
  • (D)Solution (B) is the correct answer. As f ′ (x) = x cosx + sinx  π π π π So, f ′  = cos + sin = 1  2 2 2 2

Questions

Q2Short answer

2 π π lim (sec x − tan x) = lim [sec( − y ) − tan ( − y )] x→ π y →0 2 2 = lim y→0 (cosec y − cot y )  1 cos y  = lim y →0  −  sin y sin y   1 − cos y  = lim y →0   sin y  LIMITS AND DERIVATIVES 229  y 1 − cos y  2 y since , sin 2 = 2 sin  2 2  2   = lim y →0 y y  sin y = 2 sin y cos y  2sin cos  2 2  2 2 = lim tan =0 →0 2 sin (2 + x) − sin(2 − x)

Q6Long answer

6 sin x + 1 = lim (as 2 sin x – 1 ≠ 0) x → sin x − 1 π π 1 + sin = 6 = –3 π sin −1 LIMITS AND DERIVATIVES 233 tan x − sin x

Q1Short answer

lim x →3 x − 3 2. x→ 1 2 x − 1 3. lim h→0 h 1 1 5 5 ( x + 2) 3 − 2 3 (1 + x) − 1 (2 + x) − (a + 2) 2

Q4Short answer

lim 5. lim 6. lim x→0 x x →1 (1 + x ) 2 − 1 x→ a x−a x4 − x x2 − 4

Q7Short answer

lim 8. lim x →1 x −1 x→ 2 3x − 2 − x + 2 x4 − 4 x7 − 2 x5 + 1 1 + x3 − 1 − x3

Q9Short answer

xlim 10. lim 11. lim → 2 x2 + 3 2 x − 8 x →1 x3 − 3x 2 + 2 x→0 x2 x 3 + 27  8 x − 3 4 x 2 + 1 lim  −

Q12Short answer

xlim →−3 x 5 + 243 13. x → 1  2 x − 1 4 x 2 − 1 x n − 2n sin 3 x

Q14Short answer

Find ‘n’, if lim = 80 , n ∈ N 15. lim x→ 2 x − 2 x → a sin 7 x sin 2 2 x 1 − cos 2 x 2sin x − sin 2 x

Q16Short answer

lim 17. lim 18. lim x → 0 sin 2 4 x x→0 x2 x→0 x3 1 − cos 6 x sin x − cos x 1 − cos mx lim lim

Q19Short answer

lim x →0 1 − cos nx 20. x→ π π 2  − x  21. x→ π π 3  4 x− 3 sin x − cos x sin 2 x + 3 x sin x − sin a

Q22Short answer

limπ π 23. lim x → 0 2 x + tan 3 x 24. lim x→ x− x→ a x− a cot 2 x − 3 2 − 1 + cos x

Q25Short answer

limπ 26. lim x → cosec x − 2 x→0 sin 2 x sin x − 2sin 3x + sin 5 x

Q27Short answer

lim x→0 x x4 − 1 x3 − k 3

Q28Short answer

If lim = lim 2 , then find the value of k. x →1 x − 1 x→ k x − k 2 Differentiate each of the functions w. r. to x in Exercises 29 to 42. x 4 + x3 + x 2 + 1  1

Q29Short answer

30.  x +  31. (3x + 5) (1 + tanx) x x LIMITS AND DERIVATIVES 241 3x + 4 x 5 − cos x

Q32Short answer

(sec x – 1) (sec x + 1) 33. 2 34. 5x − 7 x + 9 sin x π x 2 cos

Q35Short answer

4 36. (ax2 + cotx) (p + q cosx) sin x a + b sin x

Q37Short answer

38. (sin x + cosx)2 39. (2x – 7)2 (3x + 5)3 c + d cos x

Q40Short answer

x2 sinx + cos2x 41. sin3x cos3x 42. 2 ax + bx + c

Q43Long answer

cos (x2 + 1) 44. cx + d 45. x 3

Q46Long answer

x cosx Evaluate each of the following limits in Exercises 47 to 53. ( x + y ) sec( x + y ) − x sec x

Q47Long answer

lim y→0 y (sin(α + β) x + sin(α − β) x + sin 2α x)

Q48Long answer

lim x→0 cos 2βx − cos 2αx ⋅x tan 3 x − tan x 1 − sin lim lim 2

Q49Long answer

x→ π4 cos  x + π  50. x→ π x x x  4 cos  cos − sin  2  4 4 | x − 4|

Q51Long answer

Show that lim does not exists x→4 x − 4  k cos x π  π − 2 x when x ≠  π

Q52Long answer

Let f (x) =  π and if limπ f ( x) = f ( ) , 3 x= x→ 2  2 2 find the value of k. x + 2 x≤ 1

Q53Long answer

Let f (x) =  2 , find ‘c’ if xlim f ( x) exists.  cx x > −1 → –1

Q54Multiple choice

lim is x →π x − π

  • (A)1
  • (B)2
  • (C)–1
  • (D)–2 x 2 cos x
Q55Multiple choice

lim is x → 0 1 − cos x 3 −3

  • (A)2
  • (B)
  • (C)
  • (D)1 2 2 (1 + x) n − 1
Q56Multiple choice

lim is x→0 x

  • (A)n
  • (B)1
  • (C)–n
  • (D)0 xm − 1
Q57Multiple choice

lim is x →1 x n − 1 m m m2

  • (A)1
  • (B)
  • (C)−
  • (D)2 n n n 1 − cos 4θ
Q58Multiple choice

lim is x→0 1 − cos 6θ LIMITS AND DERIVATIVES 243 4 1 −1

  • (A)
  • (B)
  • (C)
  • (D)–1 9 2 2 cosec x − cot x
Q59Multiple choice

lim is x→0 x −1 1

  • (A)
  • (B)1
  • (C)
  • (D)1 2 2 sin x
Q60Multiple choice

lim is x→0 x +1 − 1− x

  • (A)2
  • (B)0
  • (C)1
  • (D)–1 sec 2 x − 2
Q61Multiple choice

lim is π tan x − 1 x→

  • (A)3
  • (B)1
  • (C)0
  • (D)2
Q62Multiple choice

lim ( x − 1) ( 2 x − 3) is x →1 2 x2 + x − 3 1 −1

  • (A)
  • (B)
  • (C)1
  • (D)None of these 10 10  sin[ x]  , [ x] ≠ 0 [ x]
Q63Multiple choice

If f (x) =  , where [.] denotes the greatest integer function ,  0 ,[ x] = 0  then lim x→0 f ( x) is equal to

  • (A)1
  • (B)0
  • (C)–1
  • (D)None of these | sin x |
Q64Multiple choice

lim is x→0 x

  • (A)1
  • (B)–1
  • (C)does not exist(D) None of these  x 2 − 1, 0 < x < 2
Q65Multiple choice

Let f (x) =  , the quadratic equation whose roots are xlim →2– f ( x) and  2 x + 3, 2 ≤ x < 3 lim f ( x) is x → 2+

  • (A)x2 – 6x + 9 = 0
  • (B)x2 – 7x + 8 = 0
  • (C)x2 – 14x + 49 = 0
  • (D)x2 – 10x + 21 = 0 tan 2 x − x
Q66Multiple choice

lim is x → 0 3 x − sin x 1 −1 1

  • (A)2
  • (B)
  • (C)
  • (D)2 2 4  1
Q67Multiple choice

Let f (x) = x – [x]; ∈ R, then f ′   is

  • (A)
  • (B)1
  • (C)0
  • (D)–1 1 dy
Q68Multiple choice

If y = x+ , then at x = 1 is x dx 1 1

  • (A)1
  • (B)
  • (C)
  • (D)0 2 2 x−4
Q69Multiple choice

If f (x) = , then f ′(1) is 2 x 5 4

  • (A)
  • (B)
  • (C)1
  • (D)0 4 5 1+ x2 dy
Q70Multiple choice

If y = , then is 1 dx 1− 2 − 4x − 4x 1 − x2 4x

  • (A)
  • (B)
  • (C)
  • (D)2 ( x 2 − 1)2 x2 − 1 4x x −1 sin x + cos x dy
Q71Multiple choice

If y = , then at x = 0 is sin x − cos x dx LIMITS AND DERIVATIVES 245

  • (A)–2
  • (B)0
  • (C)
  • (D)does not exist sin( x + 9) dy
Q72Multiple choice

If y = , then at x = 0 is cos x dx

  • (A)cos 9
  • (B)sin 9
  • (C)0
  • (D)1 x2 x100
Q73Multiple choice

If f (x) = 1 + x + + ... + , then f ′(1) is equal to 2 100

  • (A)
  • (B)100
  • (C)does not exist
  • (D)0 xn − an
Q74Multiple choice

If f ( x ) = for some constant ‘a’, then f ′(a) is x−a

  • (A)1
  • (B)0
  • (C)does not exist
  • (D)
Q75Multiple choice

If f (x) = x100 + x99 + ... + x + 1, then f ′(1) is equal to

  • (A)5050
  • (B)5049
  • (C)5051
  • (D)50051 2 3 99 100
Q76Multiple choice

If f (x) = 1 – x + x – x ... – x + x , then f ′(1) is euqal to

  • (A)150
  • (B)–50
  • (C)–150
  • (D)50 Fill in the blanks in Exercises 77 to 80. tan x
Q77Fill in the blanks

If f (x) = , then lim x →π f ( x) = ______________ x−π  x 

Q78Fill in the blanks

lim  sin mx cot  = 2 , then m = ______________ x→0  3 x x 2 x3 dy

Q79Fill in the blanks

if y = 1 + + + + ... , then = ______________ 1! 2! 3! dx

Q80Fill in the blanks

lim+ = ______________ x→3 [ x]