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lim-x-0-tan-2x-pi-4-2tan-x-pi-4-tan-pi-4-sin-2x-pi-4-2sin-x-pi-4-sin-pi-4-




Question Number 132240 by bemath last updated on 12/Feb/21
 lim_(x→0)  ((tan (2x+(π/4))−2tan (x+(π/4))+tan (π/4))/(sin (2x+(π/4))−2sin (x+(π/4))+sin (π/4))) ?
limx0tan(2x+π4)2tan(x+π4)+tanπ4sin(2x+π4)2sin(x+π4)+sinπ4?
Answered by EDWIN88 last updated on 13/Feb/21
 lim_(x→0)  ((2sec^2 (2x+(π/4))−2sec^2 (x+(π/4)))/(2cos (2x+(π/4))−2cos (x+(π/4))))  = lim_(x→0) ((cos^2 (x+(π/4))−cos^2 (2x+(π/4)))/([ cos (2x+(π/4)).cos (x+(π/4))]^2 [cos (2x+(π/4))−cos (x+(π/4))]))  = lim_(x→0)  ((−[ cos (x+(π/4))+cos (2x+(π/4)) ])/([ cos (2x+(π/4)).cos (x+(π/4))]^2 ))  = −((√2)/(((1/2))^2 )) = −4(√2)
limx02sec2(2x+π4)2sec2(x+π4)2cos(2x+π4)2cos(x+π4)=limx0cos2(x+π4)cos2(2x+π4)[cos(2x+π4).cos(x+π4)]2[cos(2x+π4)cos(x+π4)]=limx0[cos(x+π4)+cos(2x+π4)][cos(2x+π4).cos(x+π4)]2=2(12)2=42

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