Limits-5

Prob: Find limit of (sin^-1x - tan^-1x)/x³ when x→0

when x → 0 both numerator and denominator become zero

So L'Hospital rule becomes applicable

L'Hospital rule

(i). If f and g are differentiable functions on (0,δ) such that
(ii) g'(x)≠ 0 or any x Є (0,δ)
(iii) lim x→0⁺ f(x) = 0 = lim x→0⁴ g(x)

(iv) lim x→0⁺ f'(x)/g'(x) = L, then lim x→0⁺ f(x)/g(x) = L


lim x→0(sin^-1x - tan^-1x)/x³ =

lim x→0 [1/√(1-x²) - 1/(1+x²)]/3x² =

1/3 lim x→0 (1/x²)[(1+x²) -√(1-x²)]/[(1+x²)*√(1-x²)] =

1/3 lim x→0 (1/x²)[(1+x²) -√(1-x²)]*[(1+x²) +√(1-x²)]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]

1/3 lim x→0 (1/x²)[(1+x²)² -(1-x²)]/ [(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=

1/3 lim x→0 (1/x²)[ 1 + 2x²+x⁴ -1 +x²]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=

1/3 lim x→0 (1/x²)[ 3x²+x⁴]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=

1/3 lim x→0 (1/x²)[ x²(3+x²)]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=

1/3 lim x→0 (3+x²)/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]= (1/3)(3/2) = 1/2