We are asked to find the minimum value of $k$ such that $\delta = \frac{\varepsilon}{k}$ satisfies the $\varepsilon$-$\delta$ definition of the given limit:

$\lim_{x \to 5} \left( 4 - \frac{x}{5} \right) = 3.$

Step 1: Express the limit in standard form

We want to show that:

$\lim_{x \to 5} f(x) = 3$ where $f(x) = 4 - \frac{x}{5}$.

This means for every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - 5| < \delta$, then $|f(x) - 3| < \varepsilon$.

Step 2: Work with the function $f(x)$

We need to manipulate $|f(x) - 3|$:

$|f(x) - 3| = \left| \left( 4 - \frac{x}{5} \right) - 3 \right| = \left| 1 - \frac{x}{5} \right|.$

Step 3: Simplify the inequality

Now we need to ensure that $|1 - \frac{x}{5}| < \varepsilon$. To simplify this inequality:

$\left| 1 - \frac{x}{5} \right| = \left| \frac{5 - x}{5} \right| = \frac{|x - 5|}{5}.$

So, the condition becomes:

$\frac{|x - 5|}{5} < \varepsilon.$

Step 4: Solve for $|x - 5|$

Multiply both sides of the inequality by 5:

$|x - 5| < 5\varepsilon.$

Step 5: Set $\delta$

We see that to satisfy the $\varepsilon$-$\delta$ definition, we must have $\delta = 5\varepsilon$. Therefore, we want:

$\delta = \frac{\varepsilon}{k}.$

Equating this to our expression for $\delta$, we get:

$\frac{\varepsilon}{k} = 5\varepsilon.$

Step 6: Solve for $k$

Dividing both sides by $\varepsilon$ (assuming $\varepsilon > 0$):

$\frac{1}{k} = 5 \quad \Rightarrow \quad k = \frac{1}{5}.$

Final Answer:

The minimum value of $k$ is $\boxed{\frac{1}{5}}$.

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Do you want any further details or have any questions?

Here are some related questions to explore:

1. What is the general form of the $\varepsilon$-$\delta$ definition of a limit?
2. How does the choice of $\delta$ depend on the function $f(x)$?
3. Can you provide an example of a limit where $\delta$ depends non-linearly on $\varepsilon$?
4. What happens if the limit involves infinity instead of a finite value?
5. How does the $\varepsilon$-$\delta$ definition extend to multivariable limits?

Tip: When using the $\varepsilon$-$\delta$ definition, always carefully manipulate the inequalities involving $\varepsilon$ to find the correct $\delta$.