1. **State the problem:** Find the limit $$\lim_{x\to 1} \frac{\sqrt{x+3} - \sqrt{2x-1}}{x - 1}$$ 2. **Check direct substitution:** Substitute $x=1$: $$\frac{\sqrt{1+3} - \sqrt{2(1)-1}}{1-1} = \frac{\sqrt{4} - \sqrt{1}}{0} = \frac{2 - 1}{0} = \frac{1}{0}\;\text{(undefined)}$$ Since division by zero occurs, use algebraic manipulation. 3. **Rationalize the numerator:** Multiply numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{x+3} - \sqrt{2x-1}}{x - 1} \times \frac{\sqrt{x+3} + \sqrt{2x-1}}{\sqrt{x+3} + \sqrt{2x-1}} = \frac{(x+3) - (2x -1)}{(x-1)(\sqrt{x+3} + \sqrt{2x-1})}$$ 4. **Simplify the numerator:** $$(x+3) - (2x -1) = x + 3 - 2x + 1 = -x + 4 = 4 - x$$ 5. **Rewrite the expression:** $$\frac{4 - x}{(x-1)(\sqrt{x+3} + \sqrt{2x-1})}$$ 6. **Factor numerator to match denominator:** Observe that $4 - x = -(x - 4)$ but this doesn't factor nicely with $x-1$. Instead, note: $$4 - x = 3 - (x -1)$$ but this is not helpful. Try writing numerator as $-(x-4)$. However, direct approach is to write numerator as: $$4 - x = -(x-4)$$ but as $x \to 1$, $x - 4 \neq x - 1$. Try another substitution. 7. **Rewrite numerator:** Note: $$4 - x = (1 - x) + 3 = -(x -1) + 3$$ So, $$\frac{4 - x}{(x-1)(\sqrt{x+3} + \sqrt{2x-1})} = \frac{-(x-1) + 3}{(x-1)(\sqrt{x+3} + \sqrt{2x-1})}$$ Split into two terms: $$= \frac{-(x-1)}{(x-1)(\sqrt{x+3} + \sqrt{2x-1})} + \frac{3}{(x-1)(\sqrt{x+3} + \sqrt{2x-1})}$$ First term simplifies to: $$\frac{-1}{\sqrt{x+3} + \sqrt{2x-1}}$$ Second term still has denominator $x-1$, which is problematic. This approach is complicated; try rewriting: $$4 - x = (1 - x) + 3 = -(x -1) + 3$$ Since $x-1 \to 0$, dominant term is $-(x-1)$. 8. **Rewrite numerator as $-(x-1)$ and use limits:** $$\frac{4 - x}{(x-1)(\sqrt{x+3} + \sqrt{2x-1})} = \frac{-(x-1) + 3}{(x-1)(\sqrt{x+3} + \sqrt{2x-1})}$$ Separate: $$= \frac{-(x-1)}{(x-1)(\sqrt{x+3} + \sqrt{2x-1})} + \frac{3}{(x-1)(\sqrt{x+3} + \sqrt{2x-1})} = \frac{-1}{\sqrt{x+3} + \sqrt{2x-1}} + \frac{3}{(x-1)(\sqrt{x+3} + \sqrt{2x-1})}$$ As $x \to 1$, the second term diverges, so this is not correct. 9. **Try a different approach:** Replace $x-1$ in the expression: $$\lim_{x\to 1} \frac{\sqrt{x+3} - \sqrt{2x-1}}{x - 1}$$ Write as a derivative definition form for function: $$f(x) = \sqrt{x+3} - \sqrt{2x-1}$$ So limit is: $$\lim_{x\to 1} \frac{f(x) - f(1)}{x - 1} = f'(1)$$ 10. **Calculate $f'(x)$:** $$f'(x) = \frac{d}{dx}(\sqrt{x+3}) - \frac{d}{dx}(\sqrt{2x-1}) = \frac{1}{2\sqrt{x+3}} - \frac{2}{2\sqrt{2x-1}} = \frac{1}{2\sqrt{x+3}} - \frac{1}{\sqrt{2x-1}}$$ 11. **Evaluate at $x=1$:** $$f'(1) = \frac{1}{2\sqrt{1+3}} - \frac{1}{\sqrt{2(1)-1}} = \frac{1}{2\sqrt{4}} - \frac{1}{\sqrt{1}} = \frac{1}{2 \times 2} - 1 = \frac{1}{4} - 1 = -\frac{3}{4}$$ **Final answer:** $$\boxed{-\frac{3}{4}}$$