Use infinite series to evaluate $\lim_{x \rightarrow \infty} (x^3 - 5x^2 + 1)^{\frac{1}{3}} - x$ Use infinite series to evaluate $\displaystyle\lim_{x \rightarrow \infty} (x^3 - 5x^2 + 1)^{\frac{1}{3}} - x$
I know the limit is $-\dfrac{5}{3}$ after rationalizing the expression, but I don't know how to prove it using Taylor series. Could someone give me any hints? I prefer hints to complete solutions.

Taylor series ... Take
$$
(x^3-5x^2+1)^{1/3} = x\left(1-\frac{5}{x} + \frac{1}{x^3}\right)^{1/3}
$$
Taylor series for $1-5z+z^3$ near $z=0$ is $1 - \frac{5}{3}z + o(z)$ as $z \to 0$. So as $x \to \infty$,
\begin{align}
(x^3-5x^2+1)^{1/3} - x &= x\left(1-\frac{5}{x} + \frac{1}{x^3}\right)^{1/3} - x
\\ &= x\left(1 - \frac{5}{3x} + o(1/x)\right) - x
= -\frac{5}{3} + o(1) \to -\frac{5}{3} .
\end{align}
