Question

JEE Mais Maths question is attached

Answer 1

we have to find the limit

$\quad\displaystyle\lim_{x\to a}\frac{(a^2+2x^2)^{1/3}-(3x^2)^{1/3}}{(3a^2+x^2)^{1/3}-(4x^2)^{1/3}}$

solution : first of all let's check that limit is in the form or not.

putting x = a ,

{(a² + 2a²)⅓ - (3a²)⅓}/{(3a² + a²)⅓ - (4a²)⅓}

= 0/0 , it is in the form of limit.

now applying L - Hospital rule to find the limit because form of limit is 0/0.

$\displaystyle\lim_{x\to a}\frac{1/3(a^2+2x^2)^{-2/3}4x-1/3(3x^2)^{-2/3}6x}{1/3(3a^2+x^2)^{-2/3}2x-1/3(4x^2)^{-2/3}8x}$

= $\displaystyle\lim_{x\to a}\frac{2x(a^2+2x^2)^{-2/3}-6x(3x^2)^{-2/3}}{2x(3a^2+x^2)^{-2/3}-8x(4x^2)^{-2/3}}$

= [4a(a² + 2a²)^(-2/3) - 6a(3a²)^(-2/3)]/[2a(3a² + a²)^(-2/3) - 8a(4a²)^(-2/3)]

= [4a(3a²)^(-2/3) - 6a(3a²)^(-2/3)]/[2a(4a²)^(-2/3) - 8a(4a²)^(-2/3) ]

= [-2a(3a²)^(-2/3)]/[-6a(4a²)^(-2/3)]

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

= 1/3 (4/3)^(2/3)

Therefore the value of given limit is 1/3 (4/3)^(2/3)