\( \mathop {\lim }\limits_{x \to 0} { { \left( { { e^x} - 1} \right)\sin 2x} \over {1 - \cos x}} = \)______ 。
举一反三
- 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: \(\lim \limits_{x \to \infty } {(1 - {1 \over {2x}})^{2x}} = e\)
- \(\lim \limits_{x \to 1} { { \sin \left( { { x^2} - 1} \right)} \over {x - 1}}{\rm{ = }}\)______ 。
- \( \lim \limits_{x \to 0} { { \sqrt {1 + x\sin x} - \cos x} \over { { {\sin }^2}{x \over 2}}} = \)______ 。
- $\int {{1 \over {3 + 5\cos x}}} dx = \left( {} \right)$ A: ${1 \over 4}\ln \left| {{{2\cos x + \sin x} \over {2\cos x - \sin x}}} \right| + C$ B: ${1 \over 4}\ln \left| {{{2\cos {x \over 2} + \sin {x \over 2}} \over {2\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ C: $\ln \left| {{{\cos {x \over 2} + \sin {x \over 2}} \over {\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ D: $\ln \left| {{{\cos x + \sin x} \over {\cos x - \sin x}}} \right| + C$
- \(\mathop {\lim }\limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}\)______。______