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: \(\lim \limits_{x \to \infty } {(1 - {1 \over {2x}})^{2x}} = e\)
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: \(\lim \limits_{x \to \infty } {(1 - {1 \over {2x}})^{2x}} = e\)
举一反三
- \( \mathop {\lim }\limits_{x \to 0} { { \left( { { e^x} - 1} \right)\sin 2x} \over {1 - \cos x}} = \)______ 。
- \(\lim \limits_{x \to 1} { { \sin \left( { { x^2} - 1} \right)} \over {x - 1}}{\rm{ = }}\)______ 。
- \( \lim \limits_{x \to {0^ + }} {\left( {\cot x} \right)^ { { 1 \over {\ln x}}}} \)=_____ ______
- \( \lim \limits_{x \to 0} {x^2}\sin {1 \over x} =\)______。______
- 将\(f(x) = {1 \over {2 - x}}\)展开成\(x \)的幂级数为（ ）。 A: \({1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n }}}}} \)，\(( - 2,2)\) B: \({1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n }}}}} \)，\(\left( { - 2,2} \right]\) C: \({1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n + 1}}}}} \)，\(( - 2,2)\) D: \({1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n + 1}}}}} \)，\(\left( { - 2,2} \right]\)