\(\mathop {\lim }\limits_{n \to \infty } {\left( {1 - \frac{1}{{2n}}} \right)^{n + 1}}\) is equal to

\(\mathop {\lim }\limits_{n \to \infty } {\left( {1 - \frac{1}{{2n}}}

1.	\(\mathop {\lim }\limits_{n \to \infty } {\left( {1 - \frac{1}{{2n}}} \right)^{n + 1}}\) is equal to
A.	\(\sqrt{e}\)
B.	\(\frac{1}{{\sqrt e }}\)
C.	e
D.	\(\frac{1}{{e }}\)
Answer» C. e

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\(\mathop {\lim }\limits_{n \to \infty } {\left( {1 - \frac{1}{{2n}}} \right)^{n + 1}}\) is equal to

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