Lim square root of x_1 = 2, x approach to 5, proof epsilon delta

To prove that:

\[
\lim_{x \to 5} \sqrt{x} = \sqrt{5},
\]

using the **ε-δ definition of a limit**, we proceed as follows:

### Recall the definition:
For every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( |x - 5| < \delta \), it follows that \( |\sqrt{x} - \sqrt{5}| < \varepsilon \).

### Proof:
We aim to express \( \delta \) in terms of \( \varepsilon \).

1. **Manipulate the inequality**:
   Start with:
   \[
   |\sqrt{x} - \sqrt{5}| < \varepsilon.
   \]
   Using the property of absolute values and radicals, we can rewrite:
   \[
   |\sqrt{x} - \sqrt{5}| = \frac{|x - 5|}{|\sqrt{x} + \sqrt{5}|}.
   \]
   So the inequality becomes:
   \[
   \frac{|x - 5|}{|\sqrt{x} + \sqrt{5}|} < \varepsilon.
   \]

2. **Bound the denominator**:
   Near \( x = 5 \), we know \( \sqrt{x} \) will stay close to \( \sqrt{5} \). Let \( \delta \) be small enough so that \( x \) remains in an interval, say \( |x - 5| < 1 \). This implies \( x \in (4, 6) \).

   For \( x \in (4, 6) \), we have:
   \[
   \sqrt{x} \in (\sqrt{4}, \sqrt{6}) = (2, \sqrt{6}).
   \]
   Therefore, \( \sqrt{x} + \sqrt{5} \geq 2 + \sqrt{5} \).

3. **Simplify the inequality**:
   Substituting the lower bound for \( |\sqrt{x} + \sqrt{5}| \):
   \[
   \frac{|x - 5|}{|\sqrt{x} + \sqrt{5}|} \leq \frac{|x - 5|}{2 + \sqrt{5}}.
   \]
   To ensure \( \frac{|x - 5|}{2 + \sqrt{5}} < \varepsilon \), it suffices to choose:
   \[
   |x - 5| < \varepsilon \cdot (2 + \sqrt{5}).
   \]

4. **Choose \( \delta \)**:
   Set:
   \[
   \delta = \varepsilon \cdot (2 + \sqrt{5}).
   \]
   Then, whenever \( |x - 5| < \delta \), we have:
   \[
   |\sqrt{x} - \sqrt{5}| < \varepsilon.
   \]

### Conclusion:
By choosing \( \delta = \varepsilon \cdot (2 + \sqrt{5}) \), we satisfy the condition for the limit. Thus:
\[
\lim_{x \to 5} \sqrt{x} = \sqrt{5}.
\]

---

Would you like further clarification or additional examples? 

### Related Questions:
1. Can you provide a similar epsilon-delta proof for \( \lim_{x \to a} \sqrt{x} = \sqrt{a} \)?
2. How does the epsilon-delta definition work for non-continuous functions?
3. Could you prove \( \lim_{x \to 5} x^2 = 25 \) using epsilon-delta?
4. How do we find bounds on \( |\sqrt{x} + \sqrt{a}| \) for a general \( a > 0 \)?
5. Is it possible to use epsilon-delta for piecewise-defined functions?

### Tip:
Always check the domain of the function before applying the epsilon-delta definition, as it’s valid only where the function is defined.

Explore the epsilon-delta proof for the limit of √x as x approaches 5. This detailed solution explains the steps using the epsilon-delta definition of limits, including bounding expressions and choosing delta. Suitable for undergraduate calculus students.

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