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.
 A: Let $1/x=h$ to find
$$\lim_{h\to0^+}\dfrac{(1-5h+h^3)^{1/3}-1^{1/3}}h$$
Now rationalize the numerator using $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
A: 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}
A: Perhaps I'm missing something, but it seems to me that the problem can be solved using intuition alone.  As $x \to \infty$, I would expect $(x^3 - 5x^2 + 1)^{(1/3)}$ to go towards $[(x - 5/3)^3]^{(1/3)}$.
Assuming that my intuition is correct, the overall limit would be
$(x - 5/3) - x = -5/3.$
Addendum
My intuition is based on the following idea.
If $(x^3 - 5x^2 + 1)^{(1/3)}$ goes toward $(x - k)$ then
two things are true:
(1) The overall limit is $(-k)$.
(2) Letting $D = (x^3 - 5x^2 + 1) - (x - k)^3$ 
then $D$ will represent a 2nd degree polynomial.
It seems clear to me that $k$ must be chosen so that the $x^2$ coefficient of $D$ is 0.
This is because as $x \to \infty$, the dominant term that affects the magnitude of $D$ will be its $x^2$ coefficient.
