# 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.

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

$$\lim_{h\to0^+}\dfrac{(1-5h+h^3)^{1/3}-1^{1/3}}h$$

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

• I think it's easier to rationalize without substitution – 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
$$(x - 5/3) - x = -5/3.$$

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.