JEE Main 2020 · 3 September · evening · Single correct · +4 / −1

lim _(x → a) (( a + 2x )^(1/3) - ( 3x )^(1/3))/(( 3a + x )^(1/3) - ( 4x )^(1/3)) (a ≠ 0) is equal to :

  1. (a) ( 2/9 )( 2/3 )^(1/3)
  2. (b) ( 2/3 )( 2/9 )^(1/3)
  3. (c) ( 2/3 )^(4/3)
  4. (d) ( 2/9 )^(4/3)
Answer and solution

Answer (b) ( 2/3 )( 2/9 )^(1/3)

Solution

Maths sir · 5 steps

Put x = a and both top and bottom vanish, so differentiate each once and then substitute x = a.

  1. Spot the form. At x = a the numerator is (3a)^(1/3) - (3a)^(1/3) = 0 and the denominator is (4a)^(1/3) - (4a)^(1/3) = 0. This is the 0/0 form, so L'Hôpital's rule applies.
  2. Differentiate the top. The derivative is 2/3(a+2x)^(-2/3) - (3x)^(-2/3). At x = a this becomes 2/3(3a)^(-2/3) - (3a)^(-2/3) = -1/3(3a)^(-2/3).
  3. Differentiate the bottom. The derivative is 1/3(3a+x)^(-2/3) - 4/3(4x)^(-2/3). At x = a this becomes 1/3(4a)^(-2/3) - 4/3(4a)^(-2/3) = -(4a)^(-2/3).

    L = (-1/3(3a)^(-2/3))/-(4a)^(-2/3) = 1/3(4/3)^(2/3)

  4. Match the option. The a cancels. Write (4/3)^(2/3) = 2^(4/3)/3^(2/3) = 2 × 2^(1/3)/9^(1/3), because 3^(2/3) = 9^(1/3). So L = 2/3(2/9)^(1/3).
  5. So. The limit is (2/3)(2/9)^(1/3), which is option (b).

The idea to keep. For a 0/0 limit made of cube roots, differentiate top and bottom and substitute. Then rewrite your answer in powers of 2 and 3 to match the options.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Limits

Asked about 110 times in JEE Main since 2020. This question is from 2020.