The given limit is:

\[
\lim_{x \to 1} \sqrt{x} - 1
\]

This is an indeterminate form \( 0 - 0 \). To solve this, we can try multiplying and dividing by the conjugate to eliminate the square root.

### Steps:

1. Multiply and divide the expression by the conjugate of \( \sqrt{x} - 1 \), which is \( \sqrt{x} + 1 \):

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

2. Simplify the numerator using the difference of squares:

\[
(\sqrt{x} - 1)(\sqrt{x} + 1) = x - 1
\]

So the expression becomes:

\[
\lim_{x \to 1} \frac{x - 1}{\sqrt{x} + 1}
\]

3. Now substitute \( x = 1 \):

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

Therefore, the limit is:

\[
\boxed{0}
\]

Would you like more details on any specific step?

Here are 5 related questions for practice:
1. What is the limit of \( \lim_{x \to 4} \sqrt{x} - 2 \)?
2. How do you solve limits involving indeterminate forms like \( 0/0 \)?
3. What is the conjugate and how is it useful in simplifying expressions?
4. Can limits be solved graphically as well as algebraically?
5. What are common types of indeterminate forms?

**Tip:** When dealing with square roots in limits, conjugates can help simplify expressions and remove the indeterminate form.

Explore how to solve the limit lim_{x → 1} √x − 1 using conjugate multiplication. This detailed guide simplifies an indeterminate form (0 - 0) by eliminating the square root. Suitable for high school students in Grades 11-12.

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