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.

  • 1
    $\begingroup$ Hint: The expression is$$x\left[(1+z)^{1/3}-1\right]$$where$$z=-\frac5{x}+\frac1{x^3}\to0$$ $\endgroup$ – Peter Foreman Aug 11 '20 at 18:46
  • $\begingroup$ What did you try? Hint: $(x^3-5x^2+1)^{1/3}=x(1-5/x+1/x^3)^{1/3}$. $\endgroup$ – Chrystomath Aug 11 '20 at 18:46
  • $\begingroup$ See the section titled Newton's generalized binomial theorem $\endgroup$ – user170231 Aug 11 '20 at 19:52

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}


Let $1/x=h$ to find


Now rationalize the numerator using $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

  • $\begingroup$ I think it's easier to rationalize without substitution $\endgroup$ – crystal_math Aug 11 '20 at 20:49

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
$(x - 5/3) - x = -5/3.$

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.


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