Limit of a cube root Given $\lim_{x \to +\infty} (\sqrt[3]{x^3+x^2+x+1}-\alpha x-\beta)=0$ for some $\alpha, \beta \in \mathbb{R}$. I have tried equating it as $\lim_{x \to +\infty}\alpha x+\beta=\lim_{x \to +\infty} (\sqrt[3]{x^3+x^2+x+1})$. Cubing both sides would give$$(\alpha x)^3+\beta^3+3\alpha^2x^2\beta+3\beta^2\alpha x=x^3+x^2+x+1$$
Hence,
$$\alpha^3=1; \beta^3=1;3\beta^2\alpha=1;3\alpha^2\beta=1$$
However, I am unsure if I am doing it right.
 A: For large $x$, it is guaranteed that $\left|\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right| < 1$, and so we may use a Binomial expansion :
$$ \left(1 + \left( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right)\right)^\frac13 = 1 + \frac13 \left( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right) + O\left( \frac{1}{x^2} \right) = 1 + \frac{1}{3x} + O\left( \frac{1}{x^2} \right).$$
Therefore,
$$\sqrt[3]{x^3+x^2+x+1} = x \ \sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\alpha x-\beta$$
$$ = x\left(1 + \frac{1}{3x} + O\left( \frac{1}{x^2} \right)\right) -\alpha x  -\beta = x(1 - \alpha) + (\frac{1}{3}  - \beta) + O\left( \frac{1}{x} \right) ,$$
which tends to $0$ as $x \to \infty \iff \alpha = 1$ and $\beta = \frac13.$
$$$$
Addendum:
To answer your question for general $f(x), g(x)$ (assuming they both $\to \infty$ as $x \to \infty$): if $f(x)$ and $g(x)$ are very close for large values of $x$, such that $ f(x)-g(x) \to 0$, then it is true that, for all $ \varepsilon > 0,$ no matter how small, $\exists N$ such that $|f(x) - g(x)| < \varepsilon$ for all $x > N$.
However, this does not imply that $\left|f^3(x) - g^3(x)\right| <  \varepsilon.$ In fact,
$$\left|f^3(x) - g^3(x)\right| = \left|f(x) - g(x)\right|\left(f^2(x) + f(x)g(x) + g^2(x)\right)$$
$$ = \left|f(x) - g(x))\right|\left((f(x) - g(x))^2 + 3f(x)g(x)\right) < \varepsilon\left(\varepsilon^2 + 3f(x)g(x)\right),$$
but $3f(x)g(x)$ is not necessarily bound by $\varepsilon.$ If, for example, $3f(x)g(x) > \frac{1}{\varepsilon},$ then there may be a significant/noticable difference between $f^3(x)$ and $g^3(x)$.
Using numbers as an example, consider $10^6$ and $10^6 + \frac{1}{10^3}.$ You might think that their cubes are very close, but in fact they differ by a whooping $ \approx 3 \times 10^9$ (!)
