If a=lim(x → 0) (√(1+√(1+x⁴))-√2)/x⁴ and b=lim _(x → 0) (sin ² x)/(√2-√(1+cos x)), then the value of a b³ is :

Solution

Maths sir · 4 steps

Both limits are 0/0 with square roots, so rationalise each one and watch a and b come out as reciprocals.

1. Rationalise twice for a. Multiply by √(1+√(1+x⁴))+√2 to get √(1+x⁴)-1 on top. Then multiply by √(1+x⁴)+1 to turn that into x⁴, which cancels the denominator.

   a=lim(x→ 0) 1/(√(1+√(1+x⁴))+√2)(√(1+x⁴)+1)=1/(2√2× 2)=1/4√2

2. Rationalise for b. Multiply by √2+√(1+cos x). The denominator becomes 2-(1+cos x)=1-cos x, and (sin² x)/(1-cos x)=1+cos x.
3. Take the limit. b=lim(x→ 0)(1+cos x)(√2+√(1+cos x))=2× 2√2=4√2.
4. So. Here ab=1, so ab³=b²=(4√2)²=32, which is option (c).

The idea to keep. For a 0/0 limit with a root minus a constant, multiply by the conjugate until the factor causing the zero cancels. Nested roots just need the trick applied once per layer.

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