Understanding the Limit Evaluation Problem

We are asked to evaluate the limit of the sequence given by the expression \({\left( {1 - \frac{1}{{2n}}} \right)^{n + 1}}\) as \(n\) approaches infinity. This type of problem falls under the category of Limit Evaluation for sequences.

Identifying the Indeterminate Form

First, let's analyze the form of the expression as \(n \to \infty\). The base is \(1 - \frac{1}{{2n}}\). As \(n \to \infty\), \(\frac{1}{{2n}} \to 0\), so the base approaches \(1 - 0 = 1\). The exponent is \(n+1\). As \(n \to \infty\), \(n+1 \to \infty\).

Therefore, the limit is of the Indeterminate Form \(1^\infty\). Standard limit evaluation techniques are needed for this form.

Applying the Exponential Limit Formula for Limit Evaluation

For limits of the form \(\mathop {\lim }\limits_{x \to a} f(x)^{g(x)}\) where \(\mathop {\lim }\limits_{x \to a} f(x) = 1\) and \(\mathop {\lim }\limits_{x \to a} g(x) = \infty\), we can use the formula:

\[ \mathop {\lim }\limits_{x \to a} f(x)^{g(x)} = e^{\mathop {\lim }\limits_{x \to a} g(x) (f(x)-1)} \] This is a crucial technique for evaluating an Exponential Limit involving an indeterminate form like \(1^\infty\). In our case, \(x\) is replaced by \(n\), \(a\) is \(\infty\), \(f(n) = 1 - \frac{1}{{2n}}\), and \(g(n) = n+1\).

Calculating the Exponent Limit

We need to evaluate the limit of the exponent:

\[ \mathop {\lim }\limits_{n \to \infty} g(n) (f(n)-1) \] Substitute the expressions for \(f(n)\) and \(g(n)\):

\[ \mathop {\lim }\limits_{n \to \infty} (n+1) \left( \left(1 - \frac{1}{{2n}}\right) - 1 \right) \] Simplify the term inside the parenthesis:

\[ \mathop {\lim }\limits_{n \to \infty} (n+1) \left( -\frac{1}{{2n}} \right) \] Now, multiply the terms:

\[ \mathop {\lim }\limits_{n \to \infty} -\frac{n+1}{2n} \] To evaluate this limit, we can divide both the numerator and the denominator by the highest power of \(n\) in the denominator, which is \(n\):

\[ \mathop {\lim }\limits_{n \to \infty} -\frac{\frac{n+1}{n}}{\frac{2n}{n}} = \mathop {\lim }\limits_{n \to \infty} -\frac{1 + \frac{1}{n}}{2} \] As \(n \to \infty\), \(\frac{1}{n} \to 0\). So the limit of the exponent is:

\[ -\frac{1 + 0}{2} = -\frac{1}{2} \] This is the limit of the exponent for our Limit Evaluation.

Final Result of the Sequence Limit Evaluation

Using the exponential limit formula, the original limit is \(e\) raised to the power of the limit we just calculated:

\[ \mathop {\lim }\limits_{n \to \infty } {\left( {1 - \frac{1}{{2n}}} \right)^{n + 1}} = e^{-\frac{1}{2}} \] We can rewrite \(e^{-\frac{1}{2}}\) as \(\frac{1}{e^{\frac{1}{2}}}\), which is equivalent to \(\frac{1}{\sqrt{e}}\).

Thus, the result of the Sequence Limit evaluation is \(\frac{1}{\sqrt{e}}\). This matches one of the provided options.