To evaluate the limit $\lim_{{x \to 5}} \frac{{x^7 - 5^7}}{{x - 5}}$, we can use L'Hôpital's Rule. This rule is applicable when the limit results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Given the function:

$f(x) = \frac{{x^7 - 5^7}}{{x - 5}}$

we first substitute $x = 5$:

$f(5) = \frac{{5^7 - 5^7}}{{5 - 5}} = \frac{0}{0}$

This is an indeterminate form, so we apply L'Hôpital's Rule, which states that:

$\lim_{{x \to c}} \frac{{f(x)}}{{g(x)}} = \lim_{{x \to c}} \frac{{f'(x)}}{{g'(x)}}$

if the limit on the right-hand side exists. Here, $f(x) = x^7 - 5^7$ and $g(x) = x - 5$.

First, we find the derivatives:

$f'(x) = \frac{d}{dx}(x^7 - 5^7) = 7x^6$

$g'(x) = \frac{d}{dx}(x - 5) = 1$

Now we apply L'Hôpital's Rule:

$\lim_{{x \to 5}} \frac{{x^7 - 5^7}}{{x - 5}} = \lim_{{x \to 5}} \frac{{7x^6}}{1} = 7 \cdot 5^6$

Calculating $5^6$:

$5^6 = 15625$

Thus:

$7 \cdot 15625 = 109375$

Therefore, the limit is:

$\boxed{109375}$

Would you like more details or have any questions? Here are some related questions you might be interested in:

1. How do you evaluate limits using L'Hôpital's Rule?
2. What are the conditions for using L'Hôpital's Rule?
3. Can you provide an example of evaluating a limit without L'Hôpital's Rule?
4. How do you find the derivative of a polynomial function?
5. What are indeterminate forms in limits?

Tip: Always check the form of a limit before applying L'Hôpital's Rule to ensure it results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.