Let $$f(x) = \begin{cases} (x-1)^{\frac{1}{2-x}}, & x > 1, x \neq 2 \\ k, & x = 2 \end{cases}$$. The value of k for which f is continuous at $$x = 2$$ is:

For the function to be continuous at $$x = 2$$, the left-hand and right-hand limits of $$f(x)$$ as $$x \to 2$$ must exist and must be equal to the value of the function at that point. Symbolically, continuity demands

$$\lim_{x \to 2} (x-1)^{\frac{1}{2-x}} \;=\; f(2) \;=\; k.$$

So we have to evaluate the limit

$$L \;=\; \lim_{x \to 2} (x-1)^{\frac{1}{2-x}}.$$

Direct substitution gives $$1^{\infty},$$ an indeterminate form. To resolve it, we first take the natural logarithm. Let

$$y \;=\; (x-1)^{\frac{1}{2-x}}.$$

Taking logarithms on both sides, we obtain

$$\ln y \;=\; \frac{\ln(x-1)}{\,2 - x\,}.$$

Now we must evaluate

$$\lim_{x \to 2} \frac{\ln(x-1)}{\,2 - x\,}.$$

As $$x \to 2,$$ the numerator $$\ln(x-1) \to \ln 1 = 0$$ and the denominator $$2 - x \to 0,$$ giving the indeterminate form $$\frac{0}{0}.$$ According to L’Hospital’s Rule (which states that if $$\lim_{x \to a} \frac{f(x)}{g(x)}$$ is of the form $$\frac{0}{0}$$ or $$\frac{\infty}{\infty},$$ then the limit equals $$\lim_{x \to a} \frac{f'(x)}{g'(x)}$$ provided this derivative limit exists), we may differentiate the numerator and denominator separately:

Numerator derivative: $$\dfrac{d}{dx}\bigl[\ln(x-1)\bigr] = \dfrac{1}{x-1}.$$
Denominator derivative: $$\dfrac{d}{dx}[\,2 - x\,] = -1.$$

Applying L’Hospital’s Rule, we get

$$\lim_{x \to 2} \frac{\ln(x-1)}{\,2 - x\,} \;=\; \lim_{x \to 2} \frac{\dfrac{1}{x-1}}{-1} \;=\; \lim_{x \to 2} \Bigl(-\frac{1}{x-1}\Bigr).$$

Now substituting $$x = 2$$ gives

$$-\frac{1}{2-1} = -1.$$

Thus,

$$\ln y = -1.$$

Exponentiating both sides with base $$e$$, we arrive at

$$y = e^{-1}.$$

Therefore,

$$L = e^{-1}.$$

Continuity requires $$k = L,$$ so

$$k = e^{-1}.$$

Hence, the correct answer is Option C.