Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202523 Apr 2025Morning ShiftMathematicsLimitsActual

_ x 2 x+3 x^2+5 x^3+7 x^4-166 x-2 =

Options

  1. A167
  2. B267
  3. C287
  4. D297

Correct answer

D. 297

Step-by-step solution

The limit is of the indeterminate form 0 0 when x = 2 , so L'Hôpital's Rule applies. Applying L'Hôpital's Rule: Differentiate the numerator and denominator separately. The derivative of the numerator x + 3x^2 + 5x^3 + 7x^4 - 166 is 1 + 6x + 15x^2 + 28x^3 . The derivative of the denominator x - 2 is 1 . The limit becomes _ x 2 (1 + 6x + 15x^2 + 28x^3) . Evaluating at x = 2 : 1 + 12 + 60 + 224 = 297 . The limit is 297 .

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