If lim_x 0 0.05em 3+ sin ⁡x+ cos ⁡x+ log_e ⁡ (1-x )3 tan^2 ⁡x= 13, then 2 - is equal to :

Question

Accepted Answer

Correct answer: C. (c) lim_x 0 0.05em 3+ sin ⁡x+ cos ⁡x+ log_e ⁡ (1-x )3 tan^2 ⁡x= 13 ⇒ lim_x 0 0.05em 3+ (x- x^33 !+ …)+ (1- x^22 !+ x^44 ! …)+(-x- x^22- x^33 …)3 tan^2 ⁡x= 13 ⇒ lim_x 0 0.05em (3+ )+ ( -1 )x+(- 12- 2)x^2+ …3x^2 x^2 ath tan^2 ⁡x= 13 ⇒ +3=0 , -1=0 and - 12- 23= 13 ⇒ =-3 , =1 ⇒2 - =2+3=5

$If lim_{x \to 0} \frac{3 + α sin x + β cos x + log _{e} ( 1 - x )}{3 tan ^{2} x} = \frac{1}{3}, then 2 α - β is equal to :$

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