Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main20199 Jan 2019Morning ShiftMathematicsLimitsActual

The value of l i m y → 0 1 + 1 + y 4 - 2 y 4

Options

  1. Aexists and equals 1 2 2
  2. Bexists and equals 1 4 2
  3. Cdoes not exist
  4. Dexists and equals 1 2 2 2 + 1

Correct answer

B. exists and equals 1 4 2

Step-by-step solution

To solve this question, we need to apply the concept of rationalization twice. Given limit is l i m y → 0 1 + 1 + y 4 - 2 y 4   (   0 0   form ) = l i m y → 0 1 + 1 + y 4 - 2 y 4 × 1 + 1 + y 4 + 2 1 + 1 + y 4 + 2 = l i m y → 0 1 + y 4 - 1 y 4 1 + 1 + y 4 + 2 × 1 + y 4 + 1 1 + y 4 + 1 = l i m y → 0 y 4 y 4 1 + 1 + y 4 + 2 1 + y 4 + 1 = l i m y → 0 1 1 + 1 + y 4   + 2 1 + y 4 + 1 = 1 4 2 .

Practice Limits on Quantrex Academy →

More from Limits

Let _ x 2 ( (x-2))(rx^2 + (p-2)x - 2p) (x-2)^2 = 5 for some r, p R . If the set of all possible values of q , such that the roots of the equation rx^2 - px + q = 0 lie in (0, 2) , 2026The value of _ x 0 ( x^2 ^2 x x^2 - ^2 x ) is: 2026Let f(x) = _ y 0 (1 - (xy)) (xy) y^3 . Then the number of solutions of the equation f(x) = x , x R is : 2026The product of all possible values of , for which _ x 0 ( 1 - ( x) (( +1)x) (( +2)x) ^2(( +1)x) ) = 2 , is: 2026If _ x 2 (x^3 - 5x^2 + ax + b) ( x-1 - 1) _e(x-1) = m , then a + b + m is equal to : 2026The value of _ x 0 _ e ( (e x) (e² x ) (e¹⁰ x ) ) e²-e^ 2 x is equal to 2026If _ x 0 e ^ ( a -1) x +2 ~b x+( c -2) e ^ -x x x- _ e (1+x) =2 , then a ²+ b ²+ c ² is equal to : 2026Let [ ] denote the greatest integer function and f(x)= _ n 1 n ³ _ k =1 ^ n [ k ² 3^ x ] . Then 12 _ j =1 ^ f( j ) is equal to _ _ _ _ . 2026 Full Limits list All JEE Main PYQs