Let's solve the given problem. The question asks for the limit:

$\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}.$

Solution:

This limit involves an indeterminate form $\frac{0}{0}$ when directly substituting $x = 0$. To resolve this, we will rationalize the numerator by multiplying and dividing by the conjugate of the numerator:

$\frac{\sqrt{1+x} - 1}{x} \cdot \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1}.$

This simplifies as:

$\frac{(\sqrt{1+x} - 1)(\sqrt{1+x} + 1)}{x (\sqrt{1+x} + 1)}.$

The numerator becomes:

$(\sqrt{1+x})^2 - 1^2 = 1+x - 1 = x.$

So the expression simplifies to:

$\frac{x}{x (\sqrt{1+x} + 1)} = \frac{1}{\sqrt{1+x} + 1}.$

Now, take the limit as $x \to 0$:

$\lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1} = \frac{1}{\sqrt{1+0} + 1} = \frac{1}{1+1} = \frac{1}{2}.$

Final Answer:

$\boxed{\frac{1}{2}} \quad \text{(Option D)}.$

Would you like a further breakdown of this process? Let me know!

Here are 5 related questions for practice:

1. How would you evaluate $\lim_{x \to 0} \frac{\sqrt{4+x} - 2}{x}$?
2. What is the general approach to solve indeterminate forms like $\frac{0}{0}$?
3. How do you rationalize a numerator containing a square root?
4. What is the limit of $\frac{\sin(x)}{x}$ as $x \to 0$?
5. Can the conjugate method be used for denominators? Provide an example.

Tip: When dealing with square root limits, rationalization is a powerful technique to eliminate the indeterminate form.