\( \lim _ { x \rightarrow + } ( \sqrt { \frac { x - 2 } { x + 5 } } ) ^ { x } = \)

To find the limit \( \lim _{ x \rightarrow + \infty } \left( \sqrt { \frac { x - 2 } { x + 5 } } \right) ^ { x } \), let's simplify the expression inside the limit first. As \( x \) approaches infinity, we can simplify: \[ \frac{x - 2}{x + 5} = \frac{x(1 - \frac{2}{x})}{x(1 + \frac{5}{x})} = \frac{1 - \frac{2}{x}}{1 + \frac{5}{x}}. \] Taking the limit as \( x \) approaches infinity, we find: \[ \lim_{x \to +\infty} \frac{x - 2}{x + 5} = \frac{1 - 0}{1 + 0} = 1. \] Thus, \[ \sqrt{\frac{x - 2}{x + 5}} \to \sqrt{1} = 1 \quad \text{as } x \to +\infty. \] Now our limit has transformed to: \[ \lim_{x \to +\infty} \left(1\right)^x. \] However, this limit can be deceptive in indicating indeterminate forms, if expressed differently, so we need to investigate further using the exponential form. Rewrite: \[ y = \left( \sqrt{ \frac{x - 2}{x + 5} } \right)^{x} = e^{x \ln \left( \sqrt{ \frac{x - 2}{x + 5}} \right)} = e^{\frac{x}{2} \ln \left( \frac{x - 2}{x + 5} \right)}. \] Now simplify \( \ln \left( \frac{x - 2}{x + 5} \right) \): \[ \ln \left( \frac{x - 2}{x + 5} \right) = \ln (1 - \frac{7}{x + 5}) \text{ (using properties of logarithms)}, \] As \( x \to \infty \), we can use the approximation \( \ln(1 - u) \sim -u \): \[ \ln \left( \frac{x - 2}{x + 5} \right) \sim -\frac{7}{x + 5}. \] Then, \[ \frac{x}{2} \ln \left( \frac{x - 2}{x + 5} \right) \sim \frac{x}{2} \left(-\frac{7}{x + 5}\right) \sim -\frac{7x}{2(x + 5)}. \] As \( x \to \infty \), this approaches: \[ \sim -\frac{7}{2}. \] Finally, substituting this back into the exponent: \[ e^{-\frac{7}{2}}. \] Thus, the final limit is: \[ \lim _{ x \rightarrow + \infty } \left( \sqrt { \frac { x - 2 } { x + 5 } } \right) ^ { x } = e^{-\frac{7}{2}}. \]