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Question Number 114754 by bemath last updated on 21/Sep/20
Without L′Hopital   lim_(x→(π/4))  ((((1/(cos^2 x)) −2tan x ))/(cos 2x)) ?
WithoutLHopitallimxπ4(1cos2x2tanx)cos2x?
Commented by bemath last updated on 21/Sep/20
gave kudos all sir
gavekudosallsir
Answered by john santu last updated on 21/Sep/20
let x = (π/4)+h , h→0  lim_(h→0)  ((sec^2 ((π/4)+h)−2tan((π/4)+h) )/(cos ((π/2)+2h))) =  lim_(h→0)  ((tan^2 ((π/4)+h)−2tan ((π/4)+h)+1)/(−sin 2h)) =  lim_(h→0)  (((1−tan ((π/4)+h))^2 )/(−sin 2h)) =  lim_(h→0)  (([1−(((1+tan h)/(1−tan h)))]^2 )/(−sin 2h)) =  lim_(h→0)  (((−2tan h)^2 )/(−sin 2h)) =lim_(h→0)  ((4 tan^2 h)/(−sin 2h)) = 0
letx=π4+h,h0limh0sec2(π4+h)2tan(π4+h)cos(π2+2h)=limh0tan2(π4+h)2tan(π4+h)+1sin2h=limh0(1tan(π4+h))2sin2h=limh0[1(1+tanh1tanh)]2sin2h=limh0(2tanh)2sin2h=limh04tan2hsin2h=0
Answered by Olaf last updated on 21/Sep/20
lim_(x→(π/4)) ((1+tan^2 x−2tanx)/(cos2x))  lim_(x→(π/4)) (((1−tan^2 x)^2 )/(cos2x))  u = (π/4)−x  lim_(u→0) (([1−tan((π/4)−u)^2 ]^2 )/(cos((π/2)−2u)))  lim_(u→0) (([1−(((tan(π/4)−tanu)/(1+tan(π/4)tanu)))^2 ]^2 )/(cos((π/2)−2u)))  lim_(u→0) (([1−(((1−tanu)/(1+tanu)))^2 ]^2 )/(sin2u))  lim_(u→0) (([1−(((1−tanu)/(1+tanu)))^2 ]^2 )/(sin2u))  lim_(u→0) (([((4tan^2 u)/((1+tanu)^2 ))]^2 )/(sin2u))  lim_(u→0) (([((4u^2 )/((1+u)^2 ))]^2 )/(2u)) = 0
limxπ41+tan2x2tanxcos2xlimxπ4(1tan2x)2cos2xu=π4xlimu0[1tan(π4u)2]2cos(π22u)limu0[1(tanπ4tanu1+tanπ4tanu)2]2cos(π22u)limu0[1(1tanu1+tanu)2]2sin2ulimu0[1(1tanu1+tanu)2]2sin2ulimu0[4tan2u(1+tanu)2]2sin2ulimu0[4u2(1+u)2]22u=0
Answered by bobhans last updated on 21/Sep/20
in other way   lim_(x→(π/4))  (((tan x−1)^2 )/(cos^2 x−sin^2 x)) = lim_(x→(π/4))  (((cos x−sin x)^2 )/(cos^2 x(cos x−sin x)(cos x+sin x)))  = lim_(x→(π/4))  ((cos x−sin x)/(cos^2 x(cos x+sin x))) = 0
inotherwaylimxπ4(tanx1)2cos2xsin2x=limxπ4(cosxsinx)2cos2x(cosxsinx)(cosx+sinx)=limxπ4cosxsinxcos2x(cosx+sinx)=0
Answered by MJS_new last updated on 21/Sep/20
(1/(cos^2  x))=1+tan^2  x  cos 2x =−1+2cos^2  x =((1−t^2 )/(1+t^2 ))  ⇒  lim_(x→(π/4))  (((1−tan x)(1+tan^2  x))/(1+tan x)) =0
1cos2x=1+tan2xcos2x=1+2cos2x=1t21+t2limxπ4(1tanx)(1+tan2x)1+tanx=0

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