Solving $\lim_{x\to \infty}(\sqrt[3]{x^3-5x^2+1}-x)$ with series? In "Problem-Solving Through Problems" from Loren C Larson , there is a problem (5-4-30) that says find the below limit with infinite series .
$$\lim_{x\to \infty}(\sqrt[3]{x^3-5x^2+1}-x)$$ this is easy limitation, but what is the idea to solving with  infinite series?
I don't have a clue. I thankful for any hint in advance.
 A: Let $u = \frac{1}{x}$
Then the problem becomes:
$$\lim_{u\to 0}\left(\sqrt[3]{\left(\frac{1}{u}\right)^3-5\left(\frac{1}{u}\right)^2+1}-\frac{1}{u}\right)$$
$$=\lim_{u\to 0}\left(\sqrt[3]{\frac{1-5u+u^3}{u^3}}-\frac{1}{u}\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}(1-5u+u^3)^{\frac{1}{3}}-\frac{1}{u}\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[(1-5u+u^3)^{\frac{1}{3}}-1\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[(1-\frac{5u}{3}+O(u^2))-1\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[-\frac{5u}{3}+O(u^2)\right]\right)$$
$$=\lim_{u\to 0}\left(-\frac{5}{3}+O(u)\right)$$
$$=-\frac{5}{3}$$
EDIT: $$=\lim_{u\to 0}\left(\frac{1}{u}\left[(1-5u+u^3)^{\frac{1}{3}}-1\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[-1+\sum_{n=0}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}({u^3-5u})^n\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[\sum_{n=1}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}({u^3-5u})^n\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[\frac{1}{3}(u^3-5u) + \sum_{n=2}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}u^n({u^2-5})^n\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{3}(u^2-5) + \sum_{n=2}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}u^{n-1}({u^2-5})^n\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{3}u^2-\frac{5}{3} + u\left[\sum_{m=0}^{\infty}{\,\dbinom{\tfrac{1}{3}}{m+2}}u^{m}({u^2-5})^{m+2}\right]\right)$$
$$=-\frac{5}{3}$$
A: In this resolution, "infinite" is not essential. What matters is to find an approximation of the expression that yields the same limit while leading to tractable computation.
The royal path to such approximation is the Taylor theorem, which establishes polynomial approximations to functions, around a chosen point.
It suffices to limit the development to a small number of terms to be conclusive about the limit. (See the explanation by @FlameTrap.)

IMO, the reference to infinite series is an amalgam with entire series, somehow related to the Taylor series.
