Question

lim log(cosx)/secx
x->90
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Answer 1

Answer:

as f(x) = [{log[1 + x + x2] + log[1 – x + x2]}/(sec x – cos x)] ∴  limx→0 f(x)  = limx→0 [{log[1 + x + x2] + log[1 – x + x2]}/{[1/(cos x)] – cos x}] = limx→0 [{log[(1 + x2)2 – x2]}/{sin x ∙ tan x}] = limx→0 [{log(1 + x2 + x4)}/{sin x ∙ tan x}]  [(0/0) form] = limx→0 [{log(1 + x2[1 + x2])}/{x2(1 + x2)}] ∙ x2(1 + x2) ∙ [1/{sin x ∙ tan x}] as  limx→0 [{log(1 + x)}/x] = 1 we get, limx→0 [{log(1 + x2[1 + x2])}/{x2(1 + x2)}] ∙ x2(1 + x2) ∙ [1/{[{(sin x)/x} ∙ {(tan x)/x}]x2}] = limx→0 [{log(1 + x2[1 + x2])}/{x2(1 + x2)}] ∙ limx→0 (1 + x2)∙ [1/{limx→0 [(sinx)/x] ∙ [(tanx)/x]}] = 1 × 1 × (1/1) = 1Read more on Sarthaks.com - https://www.sarthaks.com/341243/log-1-x-x-2-log-1-x-x-2-secx-cosx-for-lim-x0

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