\( \lim _{x \rightarrow \infty } (\frac{2\times -1}{2\times +1})^{x} \)

To evaluate the limit \[ \lim_{x \rightarrow \infty} \left( \frac{2x - 1}{2x + 1} \right)^{x}, \] we first simplify the expression inside the limit. We can rewrite the fraction: \[ \frac{2x - 1}{2x + 1} = \frac{2x(1 - \frac{1}{2x})}{2x(1 + \frac{1}{2x})} = \frac{1 - \frac{1}{2x}}{1 + \frac{1}{2x}}. \] As \( x \) approaches infinity, both \( \frac{1}{2x} \) terms approach 0. Therefore, we have: \[ \frac{1 - \frac{1}{2x}}{1 + \frac{1}{2x}} \rightarrow \frac{1 - 0}{1 + 0} = 1. \] Next, we need to analyze the limit of the entire expression as \( x \) approaches infinity. We can express the limit as: \[ \lim_{x \rightarrow \infty} \left( \frac{1 - \frac{1}{2x}}{1 + \frac{1}{2x}} \right)^{x}. \] To find this limit, we can take the natural logarithm: \[ \ln L = \lim_{x \rightarrow \infty} x \ln \left( \frac{1 - \frac{1}{2x}}{1 + \frac{1}{2x}} \right). \] Using the property of logarithms, we can rewrite this as: \[ \ln L = \lim_{x \rightarrow \infty} x \left( \ln(1 - \frac{1}{2x}) - \ln(1 + \frac{1}{2x}) \right). \] Using the Taylor expansion for \( \ln(1 + u) \) around \( u = 0 \): \[ \ln(1 - u) \approx -u \quad \text{and} \quad \ln(1 + u) \approx u, \] we can substitute \( u = \frac{1}{2x} \): \[ \ln(1 - \frac{1}{2x}) \approx -\frac{1}{2x} \quad \text{and} \quad \ln(1 + \frac{1}{2x}) \approx \frac{1}{2x}. \] Thus, we have: \[ \ln(1 - \frac{1}{2x}) - \ln(1 + \frac{1}{2x}) \approx -\frac{1}{2x} - \frac{1}{2x} = -\frac{1}{x}. \] Now substituting this back into our limit: \[ \ln L = \lim_{x \rightarrow \infty} x \left( -\frac{1}{x} \right) = \lim_{x \rightarrow \infty} -1 = -1. \] Exponentiating both sides gives: \[ L = e^{-1} = \frac{1}{e}. \] Thus, the final result is: \[ \lim_{x \rightarrow \infty} \left( \frac{2x - 1}{2x + 1} \right)^{x} = \frac{1}{e}. \] Therefore, the answer is \[ \boxed{\frac{1}{e}}. \]