# Limit of algebraic function $\ \lim_{x\to\infty} \sqrt[5]{x^5 - 3x^4 + 17} - x$

How to solve this limit? $$\lim_{x \to \infty}{\sqrt[5]{x^5 - 3x^4 + 17} - x}$$

-
To paraphrase Tom Leher, "Conjugate, conjugate, conjugate". Except, instead of using $(a-b)(a+b)=a^2-b^2$, use $(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4) = a^5-b^5$. – Arturo Magidin Feb 15 '11 at 4:02
Another option is to use a first-degree approximation, namely $f(x + \epsilon) \approx f(x) + \epsilon f'(x)$ (don't forget to bound the difference using Taylor's remainder term). – Yuval Filmus Feb 15 '11 at 4:07
Another option is to apply l'Hôpital's rule to $\displaystyle{\lim_{x\to\infty}\frac{\sqrt[5]{1-\frac{3}{x}+\frac{17}{x^5}}-1}{‌​\frac{1}{x}} }$. – Jonas Meyer Feb 15 '11 at 4:15
Please avoid subjective (and non-descriptive) titles like "Difficult Limit". – Aryabhata Feb 15 '11 at 5:18
@Moron : thanks , +1 from me – Arjang Feb 15 '11 at 5:46

If you want to solve ab initio, then the way to go about is the way Arturo suggested.

Note that $a^5 - b^5 = (a-b)(a^4 + ba^3+ b^2a^2 + b^3a + b^4)$ and hence $$(a-b)= \frac{a^5 - b^5}{a^4 + ba^3+ b^2a^2 + b^3a + b^4}$$

Now take $a=\sqrt[5]{x^5-3x^4+17}$ and $b=x$

$\sqrt[5]{x^5-3x^4+17}-x = \frac{(x^5-3x^4+17)-x^5}{\left(\sqrt[5] {x^5-3x^4+17} \right)^4 + x \left(\sqrt[5] {x^5-3x^4+17} \right)^3 + x^2 \left(\sqrt[5] {x^5-3x^4+17}\right)^2 + x^3 \left(\sqrt[5] {x^5-3x^4+17}\right)+ x^4}$

Simplifying, we get,

$\sqrt[5]{x^5-3x^4+17}-x = \frac{-3x^4+17}{x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^4 + x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^3 + x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^2 + x^4 \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right) + x^4}$

$\sqrt[5]{x^5-3x^4+17}-x = \frac{-3+\frac{17}{x^4}}{\left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^4 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^3 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^2 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right) + 1}$

Hence, we have $$\sqrt[5]{x^5-3x^4+17}-x = \frac{Nr(x)}{Dr(x)}$$ where $Nr(x) = -3+\frac{17}{x^4}$ and $Dr(x) = \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^4 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^3 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right)^2 + \left(\sqrt[5] {1-\frac{3}{x}+\frac{17}{x^5}} \right) + 1$

$\displaystyle \lim_{x \rightarrow \infty} Nr(x) = -3$ and $\displaystyle \lim_{x \rightarrow \infty} Dr(x) = 5$.

Hence, we have $\displaystyle \lim_{x \rightarrow \infty} Nr(x)$ and $\displaystyle \lim_{x \rightarrow \infty} Dr(x)$ exists as a real number and hence $$\displaystyle \lim_{x \rightarrow \infty} \sqrt[5]{x^5-3x^4+17}-x = \frac{\displaystyle \lim_{x \rightarrow \infty} Nr(x)}{\displaystyle \lim_{x \rightarrow \infty} Dr(x)} = \frac{-3}{5} = - \frac{3}{5}$$

-

HINT $\$ It's simply a first derivative: $\:$ changing variables $\rm\ x\to 1/x\$ yields

$$\rm\displaystyle\ \lim_{x\to\ 0^{+}}\ \frac{f(x)-f(0)}x \ \ \ for\ \ \ f(x)\ =\ (17\ x^5-3\ x+1)^{1/5}$$

Now it is very easy to calculate $\rm\ f\:'(0)\ =\: -3/5\:$.

-

I suggest writing $(x^5 - 3x^4 + 17)^{1 \over 5}$ as $x(1 - {3 \over x} + {17 \over x^5})^{1 \over 5}$ and then using that $(1 - \epsilon)^{1 \over n} = 1 - {\epsilon \over n} + O(\epsilon^2)$.

-

Note that $$x^5 - 3x^4 + 17 = \bigg(x - \frac{3}{5}\bigg)^5 + O(x^3 ),$$ and consider the mean value theorem.

EDIT: Specifically, show that $$\sqrt[5]{{\bigg(x - \frac{3}{5}\bigg)^5 + O(x^3 )}} - \sqrt[5]{{\bigg(x - \frac{3}{5}\bigg)^5 }} \to 0.$$

-