\( \lim _ { x \rightarrow 1 ^ { + } } ( 12 + \sqrt { x - 1 } e ^ { \cos ( \frac { \pi } { x - 1 } ) } ) = \)

To evaluate the limit \[ \lim_{x \rightarrow 1^+} \left( 12 + \sqrt{x - 1} e^{\cos\left(\frac{\pi}{x - 1}\right)} \right), \] we start by analyzing the behavior of the terms as \( x \) approaches \( 1 \) from the right. 1. **Behavior of \( \sqrt{x - 1} \)**: As \( x \) approaches \( 1 \) from the right, \( x - 1 \) approaches \( 0 \) and thus \( \sqrt{x - 1} \) also approaches \( 0 \). 2. **Behavior of \( \cos\left(\frac{\pi}{x - 1}\right) \)**: As \( x \) approaches \( 1 \) from the right, \( x - 1 \) approaches \( 0 \), which means \( \frac{\pi}{x - 1} \) approaches \( +\infty \). The cosine function oscillates between \(-1\) and \(1\) as its argument approaches infinity. Therefore, \( \cos\left(\frac{\pi}{x - 1}\right) \) does not converge to a single value but oscillates. 3. **Behavior of \( e^{\cos\left(\frac{\pi}{x - 1}\right)} \)**: Since \( \cos\left(\frac{\pi}{x - 1}\right) \) oscillates between \(-1\) and \(1\), \( e^{\cos\left(\frac{\pi}{x - 1}\right)} \) oscillates between \( e^{-1} \) and \( e^{1} \). Now, we can combine these observations: \[ \sqrt{x - 1} e^{\cos\left(\frac{\pi}{x - 1}\right)} \text{ approaches } 0 \cdot \text{(bounded oscillation)}. \] Since \( \sqrt{x - 1} \) approaches \( 0 \) and \( e^{\cos\left(\frac{\pi}{x - 1}\right)} \) remains bounded, the product \( \sqrt{x - 1} e^{\cos\left(\frac{\pi}{x - 1}\right)} \) approaches \( 0 \). Thus, we can conclude: \[ \lim_{x \rightarrow 1^+} \sqrt{x - 1} e^{\cos\left(\frac{\pi}{x - 1}\right)} = 0. \] Finally, substituting this result back into the limit expression, we have: \[ \lim_{x \rightarrow 1^+} \left( 12 + \sqrt{x - 1} e^{\cos\left(\frac{\pi}{x - 1}\right)} \right) = 12 + 0 = 12. \] Therefore, the limit is \[ \boxed{12}. \]