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The given limit expression is:

\[
\lim_{{x \to 13}} \frac{{4x - 13}}{{x - 13}}
\]

To solve this, let's substitute \(x = 13\) directly into the function:

\[
\frac{{4(13) - 13}}{{13 - 13}} = \frac{{52 - 13}}{{0}} = \frac{{39}}{{0}}
\]

Since dividing by zero is undefined, this limit leads to an indeterminate form. To analyze it further, we can apply **L'Hôpital's Rule**, which is useful when a limit results in an indeterminate form like \(\frac{0}{0}\).

### Applying L'Hôpital's Rule:

1. Take the derivative of the numerator: \(\frac{d}{dx}(4x - 13) = 4\).
2. Take the derivative of the denominator: \(\frac{d}{dx}(x - 13) = 1\).

Now, the limit becomes:

\[
\lim_{{x \to 13}} \frac{4}{1} = 4
\]

Thus, the value of the limit is **4**.

Would you like more details on L'Hôpital's Rule or have any questions?

### Related Questions:
1. What is L'Hôpital's Rule, and when can it be applied?
2. How can we identify an indeterminate form like \(\frac{0}{0}\)?
3. What happens if both the numerator and denominator have higher powers of \(x\)?
4. How does substitution work in limit problems?
5. What is the difference between a finite and infinite limit?

**Tip:** When encountering a \(\frac{0}{0}\) indeterminate form in limits, always consider applying L'Hôpital's Rule by differentiating the numerator and denominator separately.

Learn how to solve the limit lim(x→13) [(4x - 13) / (x - 13)] using L'Hôpital's Rule. This problem covers key concepts of limits, indeterminate forms, and applying differentiation in limits. Suitable for advanced high school or college students.

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