Limit Root Fraction 812083

1. **State the problem:** We want to find the limit $$\lim_{x \to 3} \frac{\sqrt{x^2 + 7}}{x + 1}$$. 2. **Recall the limit rules:** If the function is continuous at the point, we can directly substitute the value of $x$. 3. **Check continuity:** The numerator $\sqrt{x^2 + 7}$ is continuous for all real $x$ because the expression inside the square root is always positive. 4. **Evaluate the denominator at $x=3$:** $3 + 1 = 4 \neq 0$, so the denominator is not zero. 5. **Direct substitution:** Substitute $x=3$ into numerator and denominator: $$\sqrt{3^2 + 7} = \sqrt{9 + 7} = \sqrt{16} = 4$$ $$3 + 1 = 4$$ 6. **Calculate the limit:** $$\lim_{x \to 3} \frac{\sqrt{x^2 + 7}}{x + 1} = \frac{4}{4} = 1$$ **Final answer:** $$1$$