1. **State the problem:** Find the one-sided limits of the function $$f(x) = \frac{\sqrt{2x}(x-1)}{|x-1|}$$ as $$x \to 1^+$$ and $$x \to 1^-$$ respectively. 2. **Rewrite the function:** The function can be expressed by considering the absolute value in the denominator: $$f(x) = \frac{\sqrt{2x}(x-1)}{|x-1|} = \sqrt{2x} \cdot \frac{x-1}{|x-1|}$$ 3. **Consider the sign of $$x - 1$$:** - For $$x > 1$$, $$x-1 > 0$$ so $$\frac{x-1}{|x-1|} = 1$$ - For $$x < 1$$, $$x-1 < 0$$ so $$\frac{x-1}{|x-1|} = -1$$ 4. **Evaluate the right-hand limit ($$x \to 1^+$$):** $$ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \sqrt{2x} \cdot 1 = \sqrt{2 \cdot 1} = \sqrt{2} $$ 5. **Evaluate the left-hand limit ($$x \to 1^-$$):** $$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \sqrt{2x} \cdot (-1) = -\sqrt{2 \cdot 1} = -\sqrt{2} $$ 6. **Final answers:** - $$\lim_{x \to 1^+} f(x) = \sqrt{2}$$ - $$\lim_{x \to 1^-} f(x) = -\sqrt{2}$$