\( \lim \limits_{x \to 0} { { x - \sin x} \over { { x^3}}} \)=( )
A: 0
B: 1
C: 6
D: \( {1 \over 6} \)
A: 0
B: 1
C: 6
D: \( {1 \over 6} \)
举一反三
- \( \lim \limits_{x \to 0} { { x - \arcsin x} \over { { {\sin }^3}x}} = {1 \over 6} \)
- \( \lim \limits_{x \to 0} {x^2}\sin {1 \over x} =\)______。______
- 求极限\( \lim \limits_{x \to 0} { { {x^2}\sin {1 \over x}} \over {\sin x}}{\rm{ = }}\)______
- \( \lim \limits_{x \to 0} { { \sqrt {1 + x\sin x} - \cos x} \over { { {\sin }^2}{x \over 2}}} = \)______ 。
- 下列极限计算正确的是( ). A: \(\lim \limits_{x \to 0} { { \left| x \right|} \over x} = 1\) B: \(\lim \limits_{x \to {0^ + }} { { \left| x \right|} \over x} = 1\) C: \(\lim \limits_{x \to 0} {(1 - {1 \over {2x}})^{2x}} = {e^{ - 1}}\) D: \(\lim \limits_{x \to \infty } {(1 - {1 \over {2x}})^{2x}} = e\)