The limit in the given expression is:

$\lim_{x \to 0} \frac{\sqrt{5 + x} - \sqrt{5}}{2x}$

To solve this limit, we can apply a technique called rationalizing the numerator. Here’s the step-by-step process:

Step 1: Multiply by the conjugate

Multiply the numerator and denominator by the conjugate of the numerator to simplify the square roots.

$\frac{\sqrt{5 + x} - \sqrt{5}}{2x} \times \frac{\sqrt{5 + x} + \sqrt{5}}{\sqrt{5 + x} + \sqrt{5}} = \frac{\left( \sqrt{5 + x} - \sqrt{5} \right) \left( \sqrt{5 + x} + \sqrt{5} \right)}{2x \left( \sqrt{5 + x} + \sqrt{5} \right)}$

Step 2: Simplify the numerator

Using the difference of squares, we get:

$\frac{(5 + x) - 5}{2x \left( \sqrt{5 + x} + \sqrt{5} \right)} = \frac{x}{2x \left( \sqrt{5 + x} + \sqrt{5} \right)}$

Step 3: Cancel the $x$

Now, cancel the $x$ in the numerator and denominator:

$\frac{1}{2 \left( \sqrt{5 + x} + \sqrt{5} \right)}$

Step 4: Take the limit as $x \to 0$

Now, substitute $x = 0$ into the remaining expression:

$\frac{1}{2 \left( \sqrt{5 + 0} + \sqrt{5} \right)} = \frac{1}{2 \left( \sqrt{5} + \sqrt{5} \right)} = \frac{1}{2 \times 2\sqrt{5}} = \frac{1}{4\sqrt{5}}$

Thus, the value of the limit is:

$\boxed{\frac{1}{4\sqrt{5}}}$

Would you like more details on this solution or have any other questions?

Related questions:

1. What are other ways to simplify limits involving square roots?
2. How do you apply the conjugate method to more complex expressions?
3. What is L'Hopital's Rule and when is it applicable?
4. Can limits always be solved algebraically, or are there cases requiring different approaches?
5. What other techniques are commonly used for indeterminate forms?

Tip: When dealing with limits that result in an indeterminate form like $\frac{0}{0}$, techniques like rationalizing or applying L'Hopital's Rule are very useful.