Let f(x) = sqrt{frac{x}{1-x}} + sqrt{frac{1-x | vedclass

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(C) Given $f(x) = \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}}$.Let $t = \sqrt{\frac{x}{1-x}}$. Then $f(x) = t + \frac{1}{t}$.We are given $\lim_{x \ri

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$\{1/5, 4/5\}$

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(C) Given $f(x) = \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}}$.Let $t = \sqrt{\frac{x}{1-x}}$. Then $f(x) = t + \frac{1}{t}$.We are given $\lim_{x \rightarrow m} (t + \frac{1}{t}) = 5/2$.This implies $t + \frac{1}{t} = 5/2$,which simplifies to $2t^2 - 5t + 2 = 0$.Solving for $t$: $(2t - 1)(t - 2) = 0$,so $t = 1/2$ or $t = 2$.Case $1$: $\sqrt{\frac{m}{1-m}} = 1/2$ $\Rightarrow \frac{m}{1-m} = 1/4$ $\Rightarrow 4m = 1 - m$ $\Rightarrow 5m = 1$ $\Rightarrow m = 1/5$.Case $2$: $\sqrt{\frac{m}{1-m}} = 2$ $\Rightarrow \frac{m}{1-m} = 4$ $\Rightarrow m = 4 - 4m$ $\Rightarrow 5m = 4$ $\Rightarrow m = 4/5$.Thus,the set of possible values for $m$ is $\{1/5, 4/5\}$.

Let $f(x) = \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}}$. If $\lim_{x \rightarrow m} f(x) = 5/2$,then the set of all possible finite values of $m$ is:

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