Solution
Related Formula
x → 0 ( x)/(x) = 1Rationalization: (u-v)(u+v) = u² - v²
Core Logic
Evaluate limit a by rationalizing the numerator:
a = x → 0 1+√(1+x⁴)-√(2)x⁴Multiply by conjugate:
a = x → 0 (1+√(1+x⁴)) - 2x⁴ ( 1+√(1+x⁴) + √(2)) a = x → 0 √(1+x⁴) - 1x⁴ ( 1+√(1+x⁴) + √(2))Rationalize again:
a = x → 0 (1+x⁴) - 1x⁴ ( 1+√(1+x⁴) + √(2)) (√(1+x⁴) + 1)Cancel x⁴:
a = x → 0 1( 1+√(1+0) + √(2)) (√(1+0) + 1) a = 1(√(2) + √(2))(1 + 1) = 14√(2)Step 1: Evaluating Limit b
Evaluate limit b by rationalizing the denominator:
b = x → 0 ² x√(2)-√(1+ x)Multiply by conjugate:
b = x → 0 ² x (√(2) + √(1+ x))2 - (1+ x) b = x → 0 (1- ² x)(√(2) + √(1+ x))1 - xUsing 1- ² x = (1- x)(1+ x):
b = x → 0 (1+ x)(√(2) + √(1+ x))Apply limit x → 0 (so 0 = 1):
b = (1+1)(√(2) + √(1+1)) = 2(2√(2)) = 4√(2)Step 2: Final Output
Calculate the value of ab³:
ab³ = ( 14√(2)) × (4√(2))³ ab³ = (4√(2))³4√(2) = (4√(2))² ab³ = 16 × 2 = 32Pattern Recognition
Double square-root structures require double rationalization. Do not rush to L'Hopital's rule when roots are stacked; iterative conjugation resolves xⁿ terms naturally.
Chapter Mix
Class 11 Maths: Limits and Derivatives Class 11 Maths: Trigonometric Functions