1. **State the problem:** We need to find the limit $$\lim_{x \to 1} \frac{\sqrt{x} - \frac{1}{\sqrt{x}}}{1-x}$$. 2. **Recall the formula and rules:** When direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, we use algebraic manipulation or L'Hôpital's Rule to evaluate the limit. 3. **Check direct substitution:** Substitute $x=1$: $$\frac{\sqrt{1} - \frac{1}{\sqrt{1}}}{1-1} = \frac{1 - 1}{0} = \frac{0}{0}$$, which is indeterminate. 4. **Manipulate the expression:** Rewrite numerator: $$\sqrt{x} - \frac{1}{\sqrt{x}} = \frac{x - 1}{\sqrt{x}}$$. 5. **Substitute back into the limit:** $$\lim_{x \to 1} \frac{\frac{x - 1}{\sqrt{x}}}{1 - x} = \lim_{x \to 1} \frac{x - 1}{\sqrt{x} (1 - x)}$$. 6. **Simplify the fraction:** Note that $1 - x = -(x - 1)$, so: $$\lim_{x \to 1} \frac{x - 1}{\sqrt{x} (1 - x)} = \lim_{x \to 1} \frac{x - 1}{\sqrt{x} \cdot -(x - 1)} = \lim_{x \to 1} \frac{\cancel{x - 1}}{\sqrt{x} \cdot -\cancel{(x - 1)}} = \lim_{x \to 1} \frac{1}{-\sqrt{x}}$$. 7. **Evaluate the limit:** Substitute $x=1$: $$\frac{1}{-\sqrt{1}} = \frac{1}{-1} = -1$$. **Final answer:** $$-1$$