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.
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: $$=\lim_{u\to 0}\left(\frac{1}{u}\left[(1-5u+u^3)^{\frac{1}{3}}-1\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[-1+\sum_{n=0}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}({u^3-5u})^n\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[\sum_{n=1}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}({u^3-5u})^n\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[\frac{1}{3}(u^3-5u) + \sum_{n=2}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}u^n({u^2-5})^n\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{3}(u^2-5) + \sum_{n=2}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}u^{n-1}({u^2-5})^n\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{3}u^2-\frac{5}{3} + u\left[\sum_{m=0}^{\infty}{\,\dbinom{\tfrac{1}{3}}{m+2}}u^{m}({u^2-5})^{m+2}\right]\right)$$
$$=-\frac{5}{3}$$