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: $\lim_{x \to 5} \sqrt{x} = \sqrt{5}.$

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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.