We are given the limit $ \lim_{x \to \infty} (\sqrt[3]{x^3 + ax^2} - \sqrt[3]{bx^3 - 5x^2 + 1}) = 2 $ and we need to find the values of $a$ and $b$.

1. Problem Description

We are given the limit

$\lim_{x \to \infty} (\sqrt[3]{x^3 + ax^2} - \sqrt[3]{bx^3 - 5x^2 + 1}) = 2$

and we need to find the values of $a$ and $b$.

2. Solution Steps

First, we factor out $x^3$ from each cube root:

$\lim_{x \to \infty} (\sqrt[3]{x^3(1 + \frac{a}{x})} - \sqrt[3]{x^3(b - \frac{5}{x} + \frac{1}{x^3})}) = 2$

$\lim_{x \to \infty} (x\sqrt[3]{1 + \frac{a}{x}} - x\sqrt[3]{b - \frac{5}{x} + \frac{1}{x^3}}) = 2$

$\lim_{x \to \infty} x(\sqrt[3]{1 + \frac{a}{x}} - \sqrt[3]{b - \frac{5}{x} + \frac{1}{x^3}}) = 2$

For the limit to be finite and non-zero, the dominant terms must cancel each other. In other words, as $x \to \infty$, $x$ must be multiplied by something that approaches $0$. This implies that $b=1$. If $b \neq 1$, the limit will either be $\infty$ or $-\infty$. So, we assume $b=1$ and proceed.

Now, let $t = 1/x$. Then as $x \to \infty$, $t \to 0$. The limit becomes

$\lim_{t \to 0} \frac{\sqrt[3]{1 + at} - \sqrt[3]{1 - 5t + t^3}}{t} = 2$

We can use the binomial approximation $(1+x)^n \approx 1 + nx$ for small $x$. Then

$\sqrt[3]{1 + at} = (1 + at)^{1/3} \approx 1 + \frac{1}{3}at$

$\sqrt[3]{1 - 5t + t^3} = (1 - 5t + t^3)^{1/3} \approx 1 + \frac{1}{3}(-5t + t^3) \approx 1 - \frac{5}{3}t$

So the limit becomes

$\lim_{t \to 0} \frac{(1 + \frac{a}{3}t) - (1 - \frac{5}{3}t)}{t} = 2$

$\lim_{t \to 0} \frac{\frac{a}{3}t + \frac{5}{3}t}{t} = 2$

$\lim_{t \to 0} \frac{(\frac{a+5}{3})t}{t} = 2$

$\frac{a+5}{3} = 2$

$a+5 = 6$

$a = 1$

Therefore, $a=1$ and $b=1$.

3. Final Answer

$a = 1$, $b = 1$