- LG a
- LG b
- LG c
- LG d
Tính các giới hạn
LG a
\[\mathop {\lim }\limits_{x \to a} \frac{{\sin x - \sin a}}{{x - a}}\]
Lời giải chi tiết:
\[\begin{array}{l}\mathop {\lim }\limits_{x \to a} \frac{{\sin x - \sin a}}{{x - a}}\\ = \mathop {\lim }\limits_{x \to a} \frac{{2\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}{{2.\frac{{x - a}}{2}}}\\ = \mathop {\lim }\limits_{x \to a} \left[ {\cos \frac{{x + a}}{2}.\frac{{\sin \frac{{x - a}}{2}}}{{\frac{{x - a}}{2}}}} \right]\\ = \mathop {\lim }\limits_{x \to a} \left[ {\cos \frac{{x + a}}{2}} \right].\mathop {\lim }\limits_{x \to a} \left[ {\frac{{\sin \frac{{x - a}}{2}}}{{\frac{{x - a}}{2}}}} \right]\\ = \cos a.1\\ = \cos a\end{array}\]
Cách khác:
Xét hàm số \[y = f\left[ x \right] = \sin x\] có:
\[\begin{array}{l}\mathop {\lim }\limits_{x \to a} \frac{{\sin x - \sin a}}{{x - a}} = \mathop {\lim }\limits_{x \to a} \frac{{f\left[ x \right] - f\left[ a \right]}}{{x - a}}\\ = f'[a]\end{array}\]
Mà \[f'\left[ x \right] = \cos x \Rightarrow f'\left[ a \right] = \cos a\]
Vậy \[\mathop {\lim }\limits_{x \to a} \frac{{\sin x - \sin a}}{{x - a}} = f'\left[ a \right] = \cos a\].
LG b
\[\mathop {\lim }\limits_{x \to 1} \left[ {1 - x} \right]\tan \frac{{\pi x}}{2}\]
Lời giải chi tiết:
\[\mathop {\lim }\limits_{x \to 1} \left[ {1 - x} \right]\tan \frac{{\pi x}}{2}\]
Đặt \[t = 1 - x\], khi \[x \to 1\] thì \[t \to 0\] ta có:
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {1 - x} \right]\tan \frac{{\pi x}}{2}\\ = \mathop {\lim }\limits_{t \to 0} \left[ {t.\tan \frac{{\pi \left[ {1 - t} \right]}}{2}} \right]\\ = \mathop {\lim }\limits_{t \to 0} \left[ {t.\tan \left[ {\frac{\pi }{2} - \frac{{\pi t}}{2}} \right]} \right]\\ = \mathop {\lim }\limits_{t \to 0} \left[ {t.\cot \frac{{\pi t}}{2}} \right]\\ = \mathop {\lim }\limits_{t \to 0} \left[ {t.\frac{{\cos \frac{{\pi t}}{2}}}{{\sin \frac{{\pi t}}{2}}}} \right]\\ = \mathop {\lim }\limits_{t \to 0} \left[ {\frac{t}{{\sin \frac{{\pi t}}{2}}}.\cos \frac{{\pi t}}{2}} \right]\\ = \mathop {\lim }\limits_{t \to 0} \left[ {\frac{{\frac{{\pi t}}{2}.\frac{2}{\pi }}}{{\sin \frac{{\pi t}}{2}}}.\cos \frac{{\pi t}}{2}} \right]\\ = \mathop {\lim }\limits_{t \to 0} \left[ {\frac{{\frac{{\pi t}}{2}}}{{\sin \frac{{\pi t}}{2}}}.\frac{2}{\pi }.\cos \frac{{\pi t}}{2}} \right]\\ = \frac{2}{\pi }.\mathop {\lim }\limits_{t \to 0} \frac{{\frac{{\pi t}}{2}}}{{\sin \frac{{\pi t}}{2}}}.\mathop {\lim }\limits_{t \to 0} \cos \frac{{\pi t}}{2}\\ = \frac{2}{\pi }.1.1\\ = \frac{2}{\pi }\end{array}\]
LG c
\[\mathop {\lim }\limits_{x \to \frac{\pi }{3}} \frac{{2{{\sin }^2}x + \sin x - 1}}{{2{{\sin }^2}x - 3\sin x + 1}}\]
Lời giải chi tiết:
\[\begin{array}{l}\mathop {\lim }\limits_{x \to \frac{\pi }{3}} \frac{{2{{\sin }^2}x + \sin x - 1}}{{2{{\sin }^2}x - 3\sin x + 1}}\\ = \frac{{2.{{\left[ {\frac{{\sqrt 3 }}{2}} \right]}^2} + \frac{{\sqrt 3 }}{2} - 1}}{{2.{{\left[ {\frac{{\sqrt 3 }}{2}} \right]}^2} - 3.\frac{{\sqrt 3 }}{2} + 1}}\\ = \frac{{\sqrt 3 + 1}}{{5 - 3\sqrt 3 }}\end{array}\]
LG d
\[\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{{\sin }^3}x}}\]
Lời giải chi tiết:
Ta có:
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{{\sin }^3}x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin x}}{{\cos x}} - \sin x}}{{{{\sin }^3}x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x - \sin x\cos x}}{{{{\sin }^3}x\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x\left[ {1 - \cos x} \right]}}{{{{\sin }^3}x\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{{{\sin }^2}x\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\frac{x}{2}}}{{4{{\sin }^2}\frac{x}{2}{{\cos }^2}\frac{x}{2}.\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{1}{{2{{\cos }^2}\frac{x}{2}.\cos x}}\\ = \frac{1}{{2.{{\cos }^2}0.\cos 0}}\\ = \frac{1}{2}\end{array}\]