The image contains a mathematical problem where a limit needs to be evaluated. I'll extract the content and solve it.

Extracted Content:

The problem appears to be:

$\lim_{{x \to 1}} \frac{{\sqrt{x + 3} - 2}}{{x - 1}}$

We need to evaluate this limit as $x$ approaches 1.

Solution:

This limit can be evaluated using the technique known as rationalizing the numerator.

1. Given limit: $\lim_{{x \to 1}} \frac{{\sqrt{x + 3} - 2}}{{x - 1}}$

2. Rationalizing the numerator: Multiply the numerator and the denominator by the conjugate of the numerator, $\sqrt{x + 3} + 2$: $\lim_{{x \to 1}} \frac{{(\sqrt{x + 3} - 2)(\sqrt{x + 3} + 2)}}{{(x - 1)(\sqrt{x + 3} + 2)}}$

3. Simplify the numerator: The numerator becomes a difference of squares: $\lim_{{x \to 1}} \frac{{(x + 3) - 4}}{{(x - 1)(\sqrt{x + 3} + 2)}} = \lim_{{x \to 1}} \frac{{x - 1}}{{(x - 1)(\sqrt{x + 3} + 2)}}$

4. Cancel out the $(x - 1)$ term in the numerator and denominator: $\lim_{{x \to 1}} \frac{1}{{\sqrt{x + 3} + 2}}$

5. Substitute $x = 1$ into the simplified expression: $\frac{1}{{\sqrt{1 + 3} + 2}} = \frac{1}{{\sqrt{4} + 2}} = \frac{1}{{2 + 2}} = \frac{1}{4}$

Final Answer:

$\lim_{{x \to 1}} \frac{{\sqrt{x + 3} - 2}}{{x - 1}} = \frac{1}{4}$

Would you like further details or have any questions?

Related Questions:

1. What other techniques can be used to evaluate limits involving square roots?
2. How do you handle indeterminate forms when calculating limits?
3. Can L'Hôpital's Rule be applied to this limit problem? Why or why not?
4. What are the key steps in rationalizing the numerator for limits?
5. How do you determine when to use substitution directly in limit problems?

Tip:

When faced with a limit problem involving square roots, rationalizing the numerator is often a helpful first step to simplify the expression.