1. **State the problem:** Find the limit $$\lim_{y \to 1} \frac{y^2 + 3}{y - 1}$$. 2. **Recall the limit concept:** The limit evaluates the behavior of the function as $y$ approaches 1. 3. **Check direct substitution:** Substitute $y=1$: $$\frac{1^2 + 3}{1 - 1} = \frac{4}{0}$$ which is undefined (division by zero). 4. **Analyze the expression:** Since direct substitution leads to division by zero, check if the numerator also approaches zero to consider factoring or simplification. 5. **Evaluate numerator at $y=1$:** $$1^2 + 3 = 4 \neq 0$$ 6. **Conclusion:** The numerator approaches 4, but denominator approaches 0, so the limit tends to infinity or negative infinity depending on the direction. 7. **Check one-sided limits:** - As $y \to 1^+$, denominator $y-1 > 0$, so fraction $\to +\infty$. - As $y \to 1^-$, denominator $y-1 < 0$, so fraction $\to -\infty$. 8. **Final answer:** The two-sided limit does not exist because the left and right limits differ. **Summary:** $$\lim_{y \to 1^+} \frac{y^2 + 3}{y - 1} = +\infty$$ $$\lim_{y \to 1^-} \frac{y^2 + 3}{y - 1} = -\infty$$ Therefore, $$\lim_{y \to 1} \frac{y^2 + 3}{y - 1} \text{ does not exist.}$$