Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202520 Apr 2025Evening ShiftMathematicsLimitsActual

_ x 0 (7^x-1 )^4 ( x k ) (1+ x^2 3 ) 4 x =3( 7)^3

Options

  1. A4( 7)⁻¹
  2. B1 4 ( 7)⁻¹
  3. C4 ( 1 7 )
  4. D1 4 7

Correct answer

A. 4( 7)⁻¹

Step-by-step solution

Evaluate the limit L = _ x 0 (7^x - 1)^4 ( x k ) (1 + x^2 3 ) 4x given that it equals 3( 7)^3 . Apply standard limits _ x 0 7^x - 1 x = 7 , _ x 0 x x = 1 , _ x 0 (1 + x) x = 1 , and _ x 0 x x = 1 . Express each component in terms of these limits: Numerator: (7^x - 1)^4 = ( 7^x - 1 x )^4 x^4 ( 7)^4 x^4 . Denominator terms: ( x k ) = (x/k) x/k x k x k , (1 + x^2/3) = (1 + x^2/3) x^2/3 x^2 3 x^2 3 , 4x = 4x 4x 4x 4x . Substitute into L : L = _ x 0 ( 7)^4 x^4 (x/k)(x^2/3)(4x) = ( 7)^4 x^4 (4x^4)/(3k) = 3k ( 7)^4 4 . Se

Practice Limits on Quantrex Academy →

More from Limits

If _ x 0 ( p 2x + 1 - 2x x + x ) = 1 then the value of ' p ' is 2026_ x 2 ( 1 - x x ) is equal to 2026The value of _ x 3 [ 1 x-3 + 9x 27-x^3 ] is: 2026_ x 0 (1 - 2x)(3 + x) x 4x is equal to: 2026If _ x 3 ( x^2 - ax - 3b x - 3 ) = 5 , then a + b = 2026If f(x) = cases x^2 - 1 & if x 2 x + 1 & if x < 2 cases , then _ x 1 f(x) + _ x 2 f(x) = 2026_ x 4 2 2 -( x+ x)^3 1- 2 x = 2025Let [x] denote the greatest integer less than or equal to x . Then _ x 2⁺ ( [x]^3 3 - [ x 3 ]^3 )= 2025 Full Limits list All MHT CET PYQs