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We are given the following limit to evaluate:

\[
\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\sin(x)}
\]

We will use the hint to multiply the numerator and denominator by the conjugate of the numerator. The conjugate of \(\sqrt{1 + x} - 1\) is \(\sqrt{1 + x} + 1\), so we proceed by multiplying both numerator and denominator by this conjugate.

\[
\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\sin(x)} \cdot \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}
\]

This simplifies as follows:

\[
= \lim_{x \to 0} \frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{\sin(x)(\sqrt{1 + x} + 1)}
\]

The numerator simplifies using the difference of squares:

\[
= \lim_{x \to 0} \frac{(1 + x) - 1^2}{\sin(x)(\sqrt{1 + x} + 1)}
= \lim_{x \to 0} \frac{x}{\sin(x)(\sqrt{1 + x} + 1)}
\]

Now, we can separate the terms:

\[
= \lim_{x \to 0} \frac{x}{\sin(x)} \cdot \frac{1}{\sqrt{1 + x} + 1}
\]

We know that \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\), so applying this limit:

\[
= 1 \cdot \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{2}
\]

Thus, the value of the limit is:

\[
\boxed{\frac{1}{2}}
\]

Let me know if you'd like further clarification or details! Here are some related questions to expand your understanding:

1. How do you determine when to use conjugates in limits?
2. Why is \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\) an important limit in calculus?
3. What other types of limits require using trigonometric identities like \(\sin(x)\)?
4. How would the solution change if the numerator had a different radical expression?
5. How can L'Hopital's Rule be used to solve this type of limit problem?

**Tip:** When solving limits with square roots or radicals, multiplying by the conjugate is often useful to simplify the expression.

Evaluate the following limit: lim(x→0) ((sqrt(1 + x) - 1) / sin(x)). Hint: begin by multiplying the numerator and denominator by the conjugate.

Learn how to evaluate the limit lim(x→0) ((sqrt(1 + x) - 1) / sin(x)) by multiplying by the conjugate. This calculus problem applies algebraic manipulation, trigonometric limits, and the key limit identity lim(x→0) (sin(x)/x) = 1. Suitable for students in Grades 11-12 or early college calculus.

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