Question

Đã trả lời bởi chuyên gia

Answer 1

Ta có: $L = \lim_{x \to \frac{π}{2}} \frac{{(\sin x + \cos x + 1)}^{2018} + 2^{2018} . (\sin x - 2)}{4 x^{3} - π^{2} x}$

$= \lim_{x \to \frac{π}{2}} \frac{[{(\sin x + \cos x + 1)}^{2018} + 2^{2018} \sin x] - 2^{2019}}{x - \frac{π}{2}} . \frac{1}{4 x (x + \frac{π}{2})} .$

Đặt $f (x) = {(\sin x + \cos x + 1)}^{2018} + 2^{2018} . \sin x .$

Khi đó $L = f^{'} (\frac{π}{2}) . \lim_{x \to \frac{π}{2}} \frac{1}{4 x (x + \frac{π}{2})} = f^{'} (\frac{π}{2}) . \frac{1}{2 π^{2}} .$

Ta có: $f^{'} (x) = 2018. {(\sin x + \cos x + 1)}^{2017} . (\cos x - \sin x) + 2^{2018} . \cos x \Rightarrow f^{'} (\frac{π}{2}) = - {2018.2}^{2017} .$

Suy ra $L = - {2018.2}^{2017} . \frac{1}{2 π^{2}} = - \frac{{1009.2}^{2017}}{π^{2}} .$

Chú ý: Cho hàm số y=f(x) thì $f^{'} (x_{0}) = \lim_{x \to x_{0}} \frac{f (x) - f (x_{0})}{x - x_{0}} .$

Chọn B

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Answer 2