\( \lim \limits_{x \to {0^ + }} { { \ln \sin 3x} \over {\ln \sin x}} = 3 \)。
错误
举一反三
- \( \lim \limits_{x \to 0} { { 9\sin x} \over { { x^3} + 3x}} = \)______ 。
- \( \lim \limits_{x \to 0} { { \sin 3x} \over x} = \)______。______
- \(\lim \limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}\)______ 。
- \(\mathop {\lim }\limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}\)______。______
- \( \lim \limits_{x \to 0} { { x - \arcsin x} \over { { {\sin }^3}x}} = {1 \over 6} \)
内容
- 0
\( \int {\cos \ln xdx} = \)( ) A: \( {x \over 2}(\cos \ln x + \sin \ln x) + C \) B: \( {x \over 2}(\cos \ln x - \sin \ln x) + C \) C: \(- {x \over 2}(\cos \ln x + \sin \ln x) + C \) D: \(- {x \over 2}(\cos \ln x - \sin \ln x) + C \)
- 1
\( \lim \limits_{x \to 0} { { x - \sin x} \over { { x^3}}} \)=( ) A: 0 B: 1 C: 6 D: \( {1 \over 6} \)
- 2
求极限\( \lim \limits_{x \to 0} { { {x^2}\sin {1 \over x}} \over {\sin x}}{\rm{ = }}\)______
- 3
\( \lim \limits_{x \to 0} { { \sqrt {1 + x\sin x} - \cos x} \over { { {\sin }^2}{x \over 2}}} = \)______ 。
- 4
\( \lim \limits_{x \to 0} {x^2}\sin {1 \over x} =\)______。______