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