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Câu hỏi: Tính $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }} \dfrac{{{\left( \sin x+\cos x+1 \right)}^{2018}}+{{2}^{2018}}.\left( \sin x-2 \right)}{4{{x}^{3}}-{{\pi }^{2}}x}.$
A. $\dfrac{{{2}^{2019}}}{{{\pi }^{2}}}.$
B. $-\dfrac{{{1009.2}^{2017}}}{{{\pi }^{2}}}.$
C. $-\dfrac{{{2}^{2018}}}{{{\pi }^{2}}}.$
D. $\dfrac{{{1009.2}^{2018}}}{{{\pi }^{2}}}.$
A. $\dfrac{{{2}^{2019}}}{{{\pi }^{2}}}.$
B. $-\dfrac{{{1009.2}^{2017}}}{{{\pi }^{2}}}.$
C. $-\dfrac{{{2}^{2018}}}{{{\pi }^{2}}}.$
D. $\dfrac{{{1009.2}^{2018}}}{{{\pi }^{2}}}.$
Ta có: $L=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }} \dfrac{{{\left( \sin x+\cos x+1 \right)}^{2018}}+{{2}^{2018}}.\left( \sin x-2 \right)}{4{{x}^{3}}-{{\pi }^{2}}x}$
$=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }} \dfrac{\left[ {{\left( \sin x+\cos x+1 \right)}^{2018}}+{{2}^{2018}}\sin x \right]-{{2}^{2019}}}{x-\dfrac{\pi }{2}}.\dfrac{1}{4x\left( x+\dfrac{\pi }{2} \right)}.$
Đặt $f\left( x \right)={{\left( \sin x+\cos x+1 \right)}^{2018}}+{{2}^{2018}}.\sin x.$
Khi đó $L={f}'\left( \dfrac{\pi }{2} \right).\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }} \dfrac{1}{4x\left( x+\dfrac{\pi }{2} \right)}={f}'\left( \dfrac{\pi }{2} \right).\dfrac{1}{2{{\pi }^{2}}}.$
Ta có: ${f}'\left( x \right)=2018.{{\left( \sin x+\cos x+1 \right)}^{2017}}.\left( \cos x-\sin x \right)+{{2}^{2018}}.\cos x\Rightarrow {f}'\left( \dfrac{\pi }{2} \right)=-{{2018.2}^{2017}}.$
Suy ra $L=-{{2018.2}^{2017}}.\dfrac{1}{2{{\pi }^{2}}}=-\dfrac{{{1009.2}^{2017}}}{{{\pi }^{2}}}.$
Chú ý: Cho hàm số $y=f\left( x \right)$ thì ${f}'\left( {{x}_{0}} \right)=\underset{x\to {{x}_{0}}}{\mathop{\lim }} \dfrac{f\left( x \right)-f\left( {{x}_{0}} \right)}{x-{{x}_{0}}}.$
$=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }} \dfrac{\left[ {{\left( \sin x+\cos x+1 \right)}^{2018}}+{{2}^{2018}}\sin x \right]-{{2}^{2019}}}{x-\dfrac{\pi }{2}}.\dfrac{1}{4x\left( x+\dfrac{\pi }{2} \right)}.$
Đặt $f\left( x \right)={{\left( \sin x+\cos x+1 \right)}^{2018}}+{{2}^{2018}}.\sin x.$
Khi đó $L={f}'\left( \dfrac{\pi }{2} \right).\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }} \dfrac{1}{4x\left( x+\dfrac{\pi }{2} \right)}={f}'\left( \dfrac{\pi }{2} \right).\dfrac{1}{2{{\pi }^{2}}}.$
Ta có: ${f}'\left( x \right)=2018.{{\left( \sin x+\cos x+1 \right)}^{2017}}.\left( \cos x-\sin x \right)+{{2}^{2018}}.\cos x\Rightarrow {f}'\left( \dfrac{\pi }{2} \right)=-{{2018.2}^{2017}}.$
Suy ra $L=-{{2018.2}^{2017}}.\dfrac{1}{2{{\pi }^{2}}}=-\dfrac{{{1009.2}^{2017}}}{{{\pi }^{2}}}.$
Chú ý: Cho hàm số $y=f\left( x \right)$ thì ${f}'\left( {{x}_{0}} \right)=\underset{x\to {{x}_{0}}}{\mathop{\lim }} \dfrac{f\left( x \right)-f\left( {{x}_{0}} \right)}{x-{{x}_{0}}}.$
Đáp án B.