We are asked to evaluate the limit: $\lim_{x \to +\infty} \frac{(2x-1)^3(x+2)^5}{x^8-1}$

1. Problem Description

We are asked to evaluate the limit:

$\lim_{x \to +\infty} \frac{(2x-1)^3(x+2)^5}{x^8-1}$

2. Solution Steps

To evaluate the limit as $x$ approaches infinity, we can focus on the highest power of $x$ in both the numerator and the denominator.

First, let's consider the numerator: $(2x-1)^3(x+2)^5$.

$(2x-1)^3 = (2x)^3 + \cdots = 8x^3 + \cdots$

$(x+2)^5 = x^5 + \cdots$

So, $(2x-1)^3(x+2)^5 = (8x^3 + \cdots)(x^5 + \cdots) = 8x^8 + \cdots$

The leading term in the numerator is $8x^8$.

Now, let's look at the denominator: $x^8 - 1$.

The leading term in the denominator is $x^8$.

Thus,

$\lim_{x \to +\infty} \frac{(2x-1)^3(x+2)^5}{x^8-1} = \lim_{x \to +\infty} \frac{8x^8 + \cdots}{x^8 - 1}$

We can divide both the numerator and the denominator by $x^8$:

$\lim_{x \to +\infty} \frac{8 + \cdots/x^8}{1 - 1/x^8}$

As $x \to +\infty$, terms like $\frac{1}{x^8}$ approach

0. $\lim_{x \to +\infty} \frac{8 + \cdots/x^8}{1 - 1/x^8} = \frac{8 + 0}{1 - 0} = \frac{8}{1} = 8$

3. Final Answer

The limit is

8.