Question

limᵧ→₀ [√{1+√(1+y⁴)} - √(2)] ÷ [y⁴] =
(A) exists and equals 1/(4√2)
(B) exists and equals 1/[2√2(√2+1)]
(C) exists and equals 1/2√2
(D) does not exist [JEE Main 2019]

Answer 1

answer : option (A) exists and equals 1/4√2

given, lim(y → 0) [√{1 + √(1 + y⁴)} - √2]/y⁴

let me check it limit is in the form or not ,

putting y = 0, [√(1 + 1) - √2]/0 = 0/0 hence limit is in the form i.e., 0/0

now rationalisation of √{1 + √(1 + y⁴)} - √2 ,

{√(1 + y⁴) - 1}/[√{1 + √(1 + y⁴)} + √2]

again, rationalize √(1 + y⁴) - 1

(1 + y⁴ - 1)/[√(1 + y⁴) + 1]

= (y⁴ )/[√(1 + y⁴) + 1]

now, lim(y→0) (y⁴)/[√(1 + y⁴) + 1][√{1 + √(1 + y⁴) + √2]y⁴

= lim(y →0) 1/[√(1 + y⁴) + 1][√{1 + √(1 + y⁴) + √2]

= 1/2(√2 + √2)

= 1/4√2

exist and equals 1/4√2