\( \lim \limits_{x \to 0} { {5 \tan 2x} \over {\sin 5x}} = \)______ 。
2
举一反三
- \(\lim \limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}\)______ 。
- \(\mathop {\lim }\limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}\)______。______
- \( \lim \limits_{x \to 0} {x^2}\sin {1 \over x} =\)______。______
- \( \lim \limits_{x \to 0} { { \sqrt {1 + x\sin x} - \cos x} \over { { {\sin }^2}{x \over 2}}} = \)______ 。
- 求极限\( \lim \limits_{x \to 0} { { {x^2}\sin {1 \over x}} \over {\sin x}}{\rm{ = }}\)______
内容
- 0
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: \(\lim \limits_{x \to \infty } {(1 - {1 \over {2x}})^{2x}} = e\)
- 1
\( \mathop {\lim }\limits_{x \to 0} { { \left( { { e^x} - 1} \right)\sin 2x} \over {1 - \cos x}} = \)______ 。
- 2
\( \lim \limits_{x \to 0} { { \sin 3x} \over x} = \)______。______
- 3
\( \lim \limits_{x \to 0} { { 9\sin x} \over { { x^3} + 3x}} = \)______ 。
- 4
\( \lim \limits_{x \to 0} { { x - \arcsin x} \over { { {\sin }^3}x}} = {1 \over 6} \)