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(B) For the function $f(x)$ to be continuous from the right at $x = 2$,the right-hand limit must equal the value of the function at $x = 2$.$\lim_{x \

Vedclass · Accepted Answer

$\frac{1}{4}$

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(B) For the function $f(x)$ to be continuous from the right at $x = 2$,the right-hand limit must equal the value of the function at $x = 2$.$\lim_{x 	o 2^+} f(x) = f(2) = k$Substituting the given function:$\lim_{x 	o 2^+} (x^2 + e^{\frac{1}{2-x}})^{-1} = k$Let $x = 2 + h$,where $h 	o 0^+$. As $h 	o 0^+$,$2 - x = 2 - (2 + h) = -h$.Thus,$\frac{1}{2-x} = -\frac{1}{h} 	o -\infty$ as $h 	o 0^+$.Therefore,$e^{\frac{1}{2-x}} = e^{-\infty} = 0$.Substituting these into the limit:$k = (2^2 + 0)^{-1} = (4)^{-1} = \frac{1}{4}$.Thus,$k = \frac{1}{4}$.

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 continuous from the right at the point $x = 2$. Then $k$ is equal to:

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