Limit of algebraic function $\ \lim_{x\to\infty} \sqrt[5]{x^5 - 3x^4 + 17} - x$ How to solve this limit?
$$\lim_{x \to \infty}{\sqrt[5]{x^5 - 3x^4 + 17} - x}$$
 A: 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 $\rm\ f\:'(0)\ =\: -3/5\:$.
A: 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 $(1 - \epsilon)^{1 \over n} =  1 - {\epsilon \over n} + O(\epsilon^2)$.
A: 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
$$\displaystyle \lim_{x \rightarrow \infty} \sqrt[5]{x^5-3x^4+17}-x = \frac{\displaystyle \lim_{x \rightarrow \infty} Nr(x)}{\displaystyle \lim_{x \rightarrow \infty} Dr(x)} = \frac{-3}{5} = - \frac{3}{5}$$
A: 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
$$
\sqrt[5]{{\bigg(x - \frac{3}{5}\bigg)^5  + O(x^3 )}} - \sqrt[5]{{\bigg(x - \frac{3}{5}\bigg)^5 }} \to 0.
$$
