For the function f(x) = frac{1}{x + 2^{frac{1 | vedclass

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(C) To determine if $f$ can be continuous at $x = 2$,we examine the limit $\lim_{x 	o 2} f(x)$.Consider the left-hand limit: $\lim_{x 	o 2^-} f(x) =

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$f$ cannot be continuous at $x = 2$

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(C) To determine if $f$ can be continuous at $x = 2$,we examine the limit $\lim_{x 	o 2} f(x)$.Consider the left-hand limit: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} \frac{1}{x + 2^{\frac{1}{x - 2}}}$.As $x 	o 2^-$,$(x - 2) 	o 0^-$,so $\frac{1}{x - 2} 	o -\infty$.Thus,$2^{\frac{1}{x - 2}} 	o 2^{-\infty} = 0$.Therefore,$\lim_{x 	o 2^-} f(x) = \frac{1}{2 + 0} = 1/2$.Now consider the right-hand limit: $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} \frac{1}{x + 2^{\frac{1}{x - 2}}}$.As $x 	o 2^+$,$(x - 2) 	o 0^+$,so $\frac{1}{x - 2} 	o +\infty$.Thus,$2^{\frac{1}{x - 2}} 	o 2^{+\infty} = \infty$.Therefore,$\lim_{x 	o 2^+} f(x) = \frac{1}{2 + \infty} = 0$.Since $\lim_{x 	o 2^-} f(x) 
eq \lim_{x 	o 2^+} f(x)$,the limit does not exist at $x = 2$.Consequently,$f$ cannot be continuous at $x = 2$ regardless of the value assigned to $f(2)$.

For the function $f(x) = \frac{1}{x + 2^{\frac{1}{x - 2}}}$,$x \neq 2$,which of the following holds?

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