We are asked to find the limit of the function $\frac{(2x-1)^3(x+2)^5}{x^8-1}$ as $x$ approaches infinity. That is, we want to find $$ \lim_{x \to \infty} \frac{(2x-1)^3(x+2)^5}{x^8-1} $$

1. Problem Description

We are asked to find the limit of the function $\frac{(2x-1)^3(x+2)^5}{x^8-1}$ as $x$ approaches infinity. That is, we want to find

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

2. Solution Steps

To evaluate the limit, we can divide both the numerator and denominator by the highest power of $x$ present, which is $x^8$. First, let's consider the highest power of $x$ in the numerator.

$(2x-1)^3$ has the highest power $x^3$, with coefficient $2^3=8$.

$(x+2)^5$ has the highest power $x^5$, with coefficient $1$.

Thus, the numerator has the highest power $x^3 \cdot x^5 = x^8$ with coefficient $8 \cdot 1 = 8$.

We have

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

Now, we divide both the numerator and denominator by $x^8$:

$$\lim_{x \to \infty} \frac{8 + \frac{\text{lower order terms}}{x^8}}{1 - \frac{1}{x^8}}$$

As $x \to \infty$, $\frac{1}{x^8} \to 0$ and $\frac{\text{lower order terms}}{x^8} \to 0$. Therefore,

$$\lim_{x \to \infty} \frac{8 + \frac{\text{lower order terms}}{x^8}}{1 - \frac{1}{x^8}} = \frac{8+0}{1-0} = \frac{8}{1} = 8$$

3. Final Answer

The final answer is

8.