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Question Number 159142 by tounghoungko last updated on 13/Nov/21

 lim_(x→−(π/4))  (((π/( (√8))) −(√2) x. tan x)/(sin x+cos x)) =?

limxπ4π82x.tanxsinx+cosx=?

Commented by cortano last updated on 14/Nov/21

 lim_(x→−(π/4)) (((π cos x−4x sin x)/( (√8)))/( (√2) sin (x+(π/4))cos x))  = (1/( 2(√2))) lim_(x→−(π/4))  ((πcos x−4x sin x)/(sin (x+(π/4))))  = (1/( 2(√2))) lim_(x→−(π/4))  ((−πsin x−4(sin x+xcos x))/(cos (x+(π/4))))  =(1/(2(√2)))((((π/( (√2)))+2(√2)+(π/( (√2))))/1))  =(1/(2(√2)))(2(√2)+π(√2))=((2+π)/2)

limxπ4πcosx4xsinx82sin(x+π4)cosx=122limxπ4πcosx4xsinxsin(x+π4)=122limxπ4πsinx4(sinx+xcosx)cos(x+π4)=122(π2+22+π21)=122(22+π2)=2+π2

Answered by FongXD last updated on 13/Nov/21

L=lim_(x→−(π/4)) (((π/( (√8)))(tanx+1)−((π/( (√8)))+(√2)x)tanx)/( (√2)(sin(π/4)sinx+cos(π/4)cosx)))  ⇔ L=(1/4)lim_(x→−(π/4)) ((((πsin(x+(π/4)))/(cosxcos(π/4)))−(π+4x)tanx)/(cos(x−(π/4))))  ⇔ L=(1/4)lim_(x→−(π/4)) ((πsin(x+(π/4))−(π+4x)tanxcosxcos(π/4))/(sin(x+(π/4))))×(1/(cosxcos(π/4)))  ⇔ L=(π/2)−(1/2)lim_(x→−(π/4)) ((π+4x)/(sin(x+(π/4))))×tanxcosxcos(π/4)  ⇔ L=(π/2)+(1/4)lim_(x→−(π/4)) ((π+4x)/(sin(x+(π/4))))  put t=x+(π/4), if x→−(π/4), ⇒ t→0  we get: L=(π/2)+(1/4)lim_(t→0) ((4t)/(sint))=(π/2)+(1/4)(4×1)=(π/2)+1  so.  determinant (((L=lim_(x→−(π/4)) (((π/( (√8)))−(√2)xtanx)/(sinx+cosx))=(π/2)+1)))

L=limxπ4π8(tanx+1)(π8+2x)tanx2(sinπ4sinx+cosπ4cosx)L=14limxπ4πsin(x+π4)cosxcosπ4(π+4x)tanxcos(xπ4)L=14limxπ4πsin(x+π4)(π+4x)tanxcosxcosπ4sin(x+π4)×1cosxcosπ4L=π212limxπ4π+4xsin(x+π4)×tanxcosxcosπ4L=π2+14limxπ4π+4xsin(x+π4)putt=x+π4,ifxπ4,t0weget:L=π2+14limt04tsint=π2+14(4×1)=π2+1so.L=limxπ4π82xtanxsinx+cosx=π2+1

Answered by tounghoungko last updated on 14/Nov/21

= (1/2) lim_(x→−(π/4))  ((π cos x−4x sin x)/( (√2) sin (x+(π/4))))  =(1/(2(√2))) lim_(x→0)  ((π cos (x−(π/4))−(4x−π)sin (x−(π/4)))/(sin x))  =(1/(2(√2))) lim_(x→0)  (((π(√2)−2(√2)x)sin x+2(√2)x cos x)/(sin x))  = (1/(2(√2))) lim_(x→0)  (π(√2)−2(√2)x+2(√2) cos x)  = ((π(√2) +2(√2))/(2(√2))) = (π/2)+1

=12limxπ4πcosx4xsinx2sin(x+π4)=122limx0πcos(xπ4)(4xπ)sin(xπ4)sinx=122limx0(π222x)sinx+22xcosxsinx=122limx0(π222x+22cosx)=π2+2222=π2+1

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