i. \( \operatorname{Lim}_{x \rightarrow 1} \frac{\sqrt{2-x}-x}{x} \)

To evaluate the limit \( \operatorname{Lim}_{x \rightarrow 1} \frac{\sqrt{2-x}-x}{x} \), first we substitute \( x = 1 \): \[ \frac{\sqrt{2-1}-1}{1} = \frac{\sqrt{1}-1}{1} = \frac{0}{1} = 0. \] Since direct substitution results in a \( \frac{0}{0} \) indeterminate form, we can apply algebraic manipulation. We multiply the numerator and denominator by the conjugate of the numerator: \[ \lim_{x \to 1} \frac{\sqrt{2-x}-x}{x} \cdot \frac{\sqrt{2-x}+x}{\sqrt{2-x}+x} = \lim_{x \to 1} \frac{(2-x)-x^2}{x(\sqrt{2-x}+x)}. \] This simplifies to: \[ \lim_{x \to 1} \frac{2 - x - x^2}{x(\sqrt{2-x}+x)} = \lim_{x \to 1} \frac{2 - 2x}{x(\sqrt{2-x}+x)} = \lim_{x \to 1} \frac{2(1-x)}{x(\sqrt{2-x}+x)}. \] Now substituting \( x = 1 \): \[ = \frac{2(1-1)}{1(\sqrt{2-1}+1)} = \frac{0}{1 + 1} = 0. \] Thus, the final answer for the limit is: \[ \operatorname{Lim}_{x \rightarrow 1} \frac{\sqrt{2-x}-x}{x} = 0. \]