Tính lim x tiến tới pi/2 (sinx+cosx+1)

* Hướng dẫn giải

Ta có: $L=\underset{x\to \frac{\pi }{2}}{\lim }\frac{{\left(\sin x+\cos x+1\right)}^{2018}+2^{2018}.\left(\sin x-2\right)}{4x^3-{\pi }^{2}x}$

$=\underset{x\to \frac{\pi }{2}}{\lim }\frac{\left[{\left(\sin x+\cos x+1\right)}^{2018}+2^{2018}\sin x\right]-2^{2019}}{x-\frac{\pi }{2}}.\frac{1}{4x\left(x+\frac{\pi }{2}\right)}.$

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

Khi đó $L=f^{\text{'}}\left(\frac{\pi }{2}\right).\underset{x\to \frac{\pi }{2}}{\lim }\frac{1}{4x\left(x+\frac{\pi }{2}\right)}=f^{\text{'}}\left(\frac{\pi }{2}\right).\frac{1}{2{\pi }^2}.$

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

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

Chú ý: Cho hàm số y=f(x) thì $f^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}.$

Chọn B