Bài 19 trang 234 SBT đại số và giải tích 11

\(\mathop {\lim }\limits_{x \to a} \frac{{\sin x - \sin a}}{{x - a}}\)

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 \(\mathop {\lim }\limits_{x \to a} \frac{{\sin x - \sin a}}{{x - a}} = f'\left( a \right) = \cos a\).

\(\mathop {\lim }\limits_{x \to 1} \left( {1 - x} \right)\tan \frac{{\pi x}}{2}\)

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:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left( {1 - x} \right)\tan \frac{{\pi x}}{2}\\ = \mathop {\lim }\limits_{t \to 0} \left[ {t.\tan \frac{{\pi \left( {1 - t} \right)}}{2}} \right]\\ = \mathop {\lim }\limits_{t \to 0} \left[ {t.\tan \left( {\frac{\pi }{2} - \frac{{\pi t}}{2}} \right)} \right]\\ = \mathop {\lim }\limits_{t \to 0} \left( {t.\cot \frac{{\pi t}}{2}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {t.\frac{{\cos \frac{{\pi t}}{2}}}{{\sin \frac{{\pi t}}{2}}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\frac{t}{{\sin \frac{{\pi t}}{2}}}.\cos \frac{{\pi t}}{2}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\frac{{\frac{{\pi t}}{2}.\frac{2}{\pi }}}{{\sin \frac{{\pi t}}{2}}}.\cos \frac{{\pi t}}{2}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\frac{{\frac{{\pi t}}{2}}}{{\sin \frac{{\pi t}}{2}}}.\frac{2}{\pi }.\cos \frac{{\pi t}}{2}} \right)\\ = \frac{2}{\pi }.\mathop {\lim }\limits_{t \to 0} \frac{{\frac{{\pi t}}{2}}}{{\sin \frac{{\pi t}}{2}}}.\mathop {\lim }\limits_{t \to 0} \cos \frac{{\pi t}}{2}\\ = \frac{2}{\pi }.1.1\\ = \frac{2}{\pi }\end{array}\)

\(\mathop {\lim }\limits_{x \to \frac{\pi }{3}} \frac{{2{{\sin }^2}x + \sin x - 1}}{{2{{\sin }^2}x - 3\sin x + 1}}\)

Lời giải chi tiết:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to \frac{\pi }{3}} \frac{{2{{\sin }^2}x + \sin x - 1}}{{2{{\sin }^2}x - 3\sin x + 1}}\\ = \frac{{2.{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2} + \frac{{\sqrt 3 }}{2} - 1}}{{2.{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2} - 3.\frac{{\sqrt 3 }}{2} + 1}}\\ = \frac{{\sqrt 3  + 1}}{{5 - 3\sqrt 3 }}\end{array}\)

\(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{{\sin }^3}x}}\)

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{{\sin }^3}x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin x}}{{\cos x}} - \sin x}}{{{{\sin }^3}x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x - \sin x\cos x}}{{{{\sin }^3}x\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x\left( {1 - \cos x} \right)}}{{{{\sin }^3}x\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{{{\sin }^2}x\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\frac{x}{2}}}{{4{{\sin }^2}\frac{x}{2}{{\cos }^2}\frac{x}{2}.\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{1}{{2{{\cos }^2}\frac{x}{2}.\cos x}}\\ = \frac{1}{{2.{{\cos }^2}0.\cos 0}}\\ = \frac{1}{2}\end{array}\)