To solve and prove the limit of the square root of $x - 1$ as $x$ approaches $5$ equals $2$ ($\lim_{x \to 5} \sqrt{x - 1} = 2$) using the $\delta$-$\epsilon$ definition, we proceed as follows:

Step 1: Recall the $\delta$-$\epsilon$ definition of a limit

The limit $\lim_{x \to 5} \sqrt{x - 1} = 2$ means that for every $\epsilon > 0$, there exists a $\delta > 0$ such that:

$0 < |x - 5| < \delta \implies |\sqrt{x - 1} - 2| < \epsilon.$

Step 2: Manipulate the inequality $|\sqrt{x - 1} - 2| < \epsilon$

We work backward to find $\delta$ in terms of $\epsilon$. Start with: $|\sqrt{x - 1} - 2| < \epsilon.$

Add $2$ to both sides: $- \epsilon < \sqrt{x - 1} - 2 < \epsilon.$

Simplify: $2 - \epsilon < \sqrt{x - 1} < 2 + \epsilon.$

Square all terms to remove the square root (valid since all terms are positive for $x > 1$): $(2 - \epsilon)^2 < x - 1 < (2 + \epsilon)^2.$

Add $1$ to all parts: $1 + (2 - \epsilon)^2 < x < 1 + (2 + \epsilon)^2.$

Step 3: Relate this to $\delta$

To ensure $|x - 5| < \delta$, we observe that: $\delta = \min\left\{5 - \left(1 + (2 - \epsilon)^2\right), \left(1 + (2 + \epsilon)^2\right) - 5\right\}.$

Step 4: Proof outline

1. Let $\epsilon > 0$ be given.
2. Define $\delta$ as the minimum distance between $x = 5$ and the bounds found in Step 3.
3. Verify that if $0 < |x - 5| < \delta$, then $|\sqrt{x - 1} - 2| < \epsilon$, completing the proof.

Would you like a detailed numeric example, or further elaboration?

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Related Questions:

1. How does the $\delta$-$\epsilon$ proof handle square root functions in general?
2. What would change in the proof if the limit point $x$ approached $a \neq 5$?
3. Can you describe why squaring is valid in this context?
4. How can we confirm the solution graphically?
5. What role does continuity of $\sqrt{x - 1}$ play in the proof?

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Tip:

When working with square roots in limits, always check the domain to avoid undefined values.