To evaluate the limit $\lim_{x \to 1} \sqrt{3 + f(x)}$ given that $\lim_{x \to 1} f(x) = 1$, we proceed as follows:

Step 1: Apply the limit to $f(x)$

Since $\lim_{x \to 1} f(x) = 1$, we can substitute this into the expression inside the square root:

$\lim_{x \to 1} \sqrt{3 + f(x)} = \sqrt{3 + \lim_{x \to 1} f(x)}$

Step 2: Substitute the known limit value

Substitute $\lim_{x \to 1} f(x) = 1$ into the equation:

$\sqrt{3 + 1} = \sqrt{4}$

Step 3: Simplify the expression

$\sqrt{4} = 2$.

Final Answer:

$\lim_{x \to 1} \sqrt{3 + f(x)} = 2$

This result holds because the square root function is continuous at the point $3 + 1 = 4$, allowing us to pass the limit through the square root.

Do you have any further questions or need more details?

Here are 5 related questions to explore further:

1. How do you prove that the square root function is continuous?
2. What are the conditions under which you can interchange limits and function operations?
3. How would the limit change if $f(x)$ approached a value different from 1?
4. What are some examples of limits where the function inside the root is not continuous?
5. How do limits behave under addition, subtraction, multiplication, and division?

Tip: When working with limits, always check the continuity of the functions involved, as it allows you to simplify the process significantly.