1. **Problem statement:** Find the limit $$\lim_{x \to 2} \frac{\sqrt{2x - 2}}{x^2 - 3x + 2}$$. 2. **Recall the formula and rules:** To find limits involving square roots and polynomials, first try direct substitution. If it results in an indeterminate form like $$\frac{0}{0}$$, use algebraic manipulation such as factoring or rationalizing. 3. **Direct substitution:** Substitute $x=2$: $$\sqrt{2(2) - 2} = \sqrt{4 - 2} = \sqrt{2} \neq 0,$$ Denominator: $$2^2 - 3(2) + 2 = 4 - 6 + 2 = 0,$$ So the denominator is zero but numerator is $$\sqrt{2}$$, so the limit tends to $$\frac{\sqrt{2}}{0}$$ which is undefined (infinite or does not exist). But let's check carefully if the numerator is zero or not. 4. **Check numerator at $x=2$ carefully:** $$2x - 2 = 2(2) - 2 = 4 - 2 = 2,$$ so numerator is $$\sqrt{2} \neq 0$$. 5. **Check denominator factorization:** $$x^2 - 3x + 2 = (x - 1)(x - 2).$$ 6. **Rewrite the limit:** $$\lim_{x \to 2} \frac{\sqrt{2x - 2}}{(x - 1)(x - 2)}.$$ 7. **Since denominator tends to zero and numerator tends to $$\sqrt{2}$$, the limit tends to infinity or negative infinity depending on the sign of denominator near 2.** 8. **Check left and right limits:** - For $$x \to 2^-$$, $$x-2 < 0$$, denominator negative, numerator positive, so limit tends to $$-\infty$$. - For $$x \to 2^+$$, $$x-2 > 0$$, denominator positive, numerator positive, so limit tends to $$+\infty$$. 9. **Conclusion:** The two-sided limit does not exist because left and right limits differ. **Final answer:** The limit does not exist. **Note:** None of the given options (1, 1/2, 2, 1/4) match this result.