Question

[{(log[1 + x + x²] + log[1 – x + x²])/(secx – cosx)}] for lim x→0 = ? 

(a) 2

(b) – 3

(c) 1

(d) – 2

Answer 1

The correct option  (c) 1

Explanation:

as f(x) = [{log[1 + x + x²] + log[1 – x + x²]}/(sec x – cos x)]

∴  lim_x→0 f(x)  = lim_x→0 [{log[1 + x + x²] + log[1 – x + x²]}/{[1/(cos x)] – cos x}]

= lim_x→0 [{log[(1 + x²)² – x²]}/{sin x ∙ tan x}]

= lim_x→0 [{log(1 + x² + x⁴)}/{sin x ∙ tan x}]  [(0/0) form]

= lim_x→0 [{log(1 + x²[1 + x²])}/{x²(1 + x²)}] ∙ x²(1 + x²) ∙ [1/{sin x ∙ tan x}]

as  lim_x→0 [{log(1 + x)}/x] = 1

we get,

lim_x→0 [{log(1 + x²[1 + x²])}/{x²(1 + x²)}] ∙ x²(1 + x²) ∙ [1/{[{(sin x)/x} ∙ {(tan x)/x}]x²}]

= lim_x→0 [{log(1 + x²[1 + x²])}/{x²(1 + x²)}] ∙ lim_x→0 (1 + x²)∙ [1/{lim_x→0 [(sinx)/x] ∙ [(tanx)/x]}]

= 1 × 1 × (1/1)

= 1