To evaluate the limit of the function $\frac{x^7 - 5^7}{x - 5}$ as $x$ approaches 5, we can use the limit properties and algebraic manipulation. The expression $\frac{x^7 - 5^7}{x - 5}$ takes the indeterminate form $\frac{0}{0}$ when $x = 5$. Therefore, we can apply L'Hôpital's Rule, which states that if $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then

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

Given: $f(x) = x^7 - 5^7$ $g(x) = x - 5$

First, we compute the derivatives $f'(x)$ and $g'(x)$:

$f'(x) = \frac{d}{dx}(x^7 - 5^7) = 7x^6$ $g'(x) = \frac{d}{dx}(x - 5) = 1$

Now, applying L'Hôpital's Rule:

[ \lim_{x \to 5} \frac{x^7 - 5^7}{x - 5} =