The image you've uploaded contains a mathematical expression. I will extract and solve it.

The expression is:

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

This is a limit problem involving a square root in the numerator and a difference of squares in the denominator. To solve this, let's go through the steps:

Step 1: Recognize the Indeterminate Form

First, substitute $x = 1$ into the expression: $\frac{{\sqrt{1 + 3} - 2}}{{1^2 - 1}} = \frac{{\sqrt{4} - 2}}{{1 - 1}} = \frac{{2 - 2}}{{0}} = \frac{0}{0}$ This is an indeterminate form, so we'll need to simplify the expression further.

Step 2: Simplify Using Algebraic Techniques

Since the numerator involves a square root, we can multiply both the numerator and the denominator by the conjugate of the numerator to eliminate the square root:

The conjugate of $\sqrt{x + 3} - 2$ is $\sqrt{x + 3} + 2$.

So, multiply the expression by $\frac{\sqrt{x + 3} + 2}{\sqrt{x + 3} + 2}$:

$\lim_{{x \to 1}} \frac{{\left(\sqrt{x + 3} - 2\right)\left(\sqrt{x + 3} + 2\right)}}{{\left(x^2 - 1\right)\left(\sqrt{x + 3} + 2\right)}}$

Step 3: Simplify the Expression Further

The numerator becomes: $(\sqrt{x + 3})^2 - 2^2 = (x + 3) - 4 = x - 1$ So the expression simplifies to: $\lim_{{x \to 1}} \frac{{x - 1}}{{(x - 1)(x + 1)\left(\sqrt{x + 3} + 2\right)}}$

Now, cancel out the $x - 1$ term from both the numerator and the denominator:

$\lim_{{x \to 1}} \frac{1}{{(x + 1)\left(\sqrt{x + 3} + 2\right)}}$

Step 4: Evaluate the Limit

Finally, substitute $x = 1$ into the simplified expression:

$\frac{1}{{(1 + 1)\left(\sqrt{1 + 3} + 2\right)}} = \frac{1}{{2 \cdot (2 + 2)}} = \frac{1}{{2 \cdot 4}} = \frac{1}{8}$

Final Answer

$\lim_{{x \to 1}} \frac{{\sqrt{x + 3} - 2}}{{x^2 - 1}} = \frac{1}{8}$

Would you like any further details or have any questions about the solution?

Related Questions:

1. How do you handle limits involving square roots when direct substitution gives an indeterminate form?
2. Can multiplying by the conjugate be applied to any indeterminate limit involving square roots?
3. What are some other algebraic techniques useful in simplifying limits?
4. How does the difference of squares in the denominator affect the limit process?
5. What would happen if the original problem had a different limit point or different function forms?

Tip:

When dealing with limits that result in the $\frac{0}{0}$ indeterminate form, look for algebraic manipulations such as factoring, conjugates, or trigonometric identities to simplify the expression before substituting the limit point.