If the function defined by f(x) = begin{cases | vedclass

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(C) Given that the function $f(x)$ is right continuous at $x = 2$. By the definition of right continuity, we have $\lim_{x 	o 2^+} f(x) = f(2)$. Here

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$\frac{1}{4}$

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(C) Given that the function $f(x)$ is right continuous at $x = 2$. By the definition of right continuity, we have $\lim_{x 	o 2^+} f(x) = f(2)$. Here, $f(2) = k$. So, $k = \lim_{h 	o 0^+} f(2+h)$. Substituting the function definition: $k = \lim_{h 	o 0^+} ((2+h)^2 + e^{\frac{1}{2-(2+h)}})^{-1}$. $k = \lim_{h 	o 0^+} ((2+h)^2 + e^{\frac{1}{-h}})^{-1}$. As $h 	o 0^+$, the term $\frac{1}{-h} 	o -\infty$, so $e^{\frac{1}{-h}} 	o e^{-\infty} = 0$. Therefore, $k = (2^2 + 0)^{-1} = (4)^{-1} = \frac{1}{4}$.

If the function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1}, & x \neq 2 \\ k, & x = 2 \end{cases}$ is right continuous at $x = 2$, then $k =$

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