Chủ đề này có 1 trả lời

[provotinhvip]

$\lim_{x\rightarrow 0}(\frac{1}{x^{2}}-(cotx)^{2})$

Bài viết đã được chỉnh sửa nội dung bởi provotinhvip: 09-11-2012 - 15:58

[vantho302]

$\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x^2}}} - {{\left( {\cot x} \right)}^2}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x^2}}} - \left( {\frac{1}{{{{\sin }^2}x}} - 1} \right)} \right) \\
= \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x^2}}} - \frac{1}{{{{\sin }^2}x}} + 1} \right) = 1 + \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x^2}}} - \frac{1}{{{{\sin }^2}x}}} \right) \\
\end{array}$

Xét $\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x^2}}} - \frac{1}{{{{\sin }^2}x}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x - {x^2}}}{{{x^2}{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x - {x^2}}}{{{x^4}}}$

Do ${\sin ^2}x \sim {x^2},x \to 0$
Dùng Lopitan ta được:
$\mathop {\lim }\limits_{x \to 0} \frac{{2\sin x\cos x - 2x}}{{4{x^3}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x - 2x}}{{4{x^3}}}$

Ta Lopitan 1 lần nữa ta được:

$\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{2\cos 2x - 2}}{{12{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{c{\rm{os}}2x - 1}}{{6{x^2}}} \\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - 2{{\sin }^2}x} \right) - 1}}{{6{x^2}}} = - \frac{1}{3}\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{{x^2}}} = - \frac{1}{3} \\
\end{array}$

Thay vào giới hạn ban đầu ta được:
$\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x^2}}} - {{\left( {\cot x} \right)}^2}} \right) = 1 - \frac{1}{3} = \frac{2}{3}$

Chúc bạn học tốt !!!