1. **State the problem:** Find the limit $$\lim_{x \to 1} \frac{\sin(x-1)}{x-1}$$. 2. **Recall the important limit formula:** The standard limit $$\lim_{t \to 0} \frac{\sin t}{t} = 1$$ is fundamental here. 3. **Substitute:** Let $$t = x - 1$$. As $$x \to 1$$, we have $$t \to 0$$. 4. **Rewrite the limit:** The limit becomes $$\lim_{t \to 0} \frac{\sin t}{t}$$. 5. **Apply the known limit:** Using the standard limit, this equals 1. 6. **Conclusion:** Therefore, $$\lim_{x \to 1} \frac{\sin(x-1)}{x-1} = 1$$. This limit evaluates to 1 because the sine function behaves like its argument near zero, making the ratio approach 1.