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.