In a math seminar while counting asymptotes, there was this limit:
$$\lim_{ x\to\infty} \sqrt[3]{3x^2 - x^3} + x$$
Its result should be $1$. (This was shown as an answer only at the seminar without a process of solution. WolframAplha says so too, but it doesn't give a step-by-step solution either.)
What is a possible step-by-step solution to this limit?
Notice that: $$ \sqrt[3]{3x^2-x^3}+x=\sqrt[3]{x^3\left(\frac3x-1\right)}+x=x\left(\sqrt[3]{\frac3x-1}+1\right)= \frac{\sqrt[3]{\frac3x-1}+1}{\frac1x}$$
This is a "$\frac00$" situtation when $x\rightarrow\infty$, so we can use L'Hôpital's rule when we pass to the limits, so:
$$\lim\limits_{x\rightarrow\infty}\left(\sqrt[3]{3x^2-x^3}+x\right)=\lim\limits_{x\rightarrow\infty}\frac{\sqrt[3]{\frac3x-1}+1}{\frac1x}=\lim\limits_{x\rightarrow\infty}\frac{\left(\sqrt[3]{\frac3x-1}+1\right)'}{\left(\frac1x\right)'}$$
$$=\lim\limits_{x\rightarrow\infty}\frac{\frac13\left(-\frac{3}{x^2}\right)\left(\frac3x-1\right)^{-2/3}}{-\frac{1}{x^2}}=\lim\limits_{x\rightarrow\infty}\left(\frac3x-1\right)^{-2/3}=\frac{1}{\sqrt[3]{(-1)^2}}=1$$
Hence the final result is: $$\lim\limits_{x\rightarrow\infty}\left(\sqrt[3]{3x^2-x^3}+x\right)=1$$
If you're not allowed to use L'Hôpital's rule, you can use Taylor's series on $\sqrt[3]{1+w}$ around $w=0$:
$$ \begin{align} &\lim_{x\to\infty} \sqrt[3]{3x^2 - x^3} + x\\ =&\lim_{x\to\infty} x+\sqrt[3]{x^3\left(\frac{3}{x} - 1\right)}\\ =&\lim_{u\to 0} \frac{1}{u}\left(\left(1 - \sqrt[3]{1-3u}\right)\right)\\ =&\lim_{u\to 0}\frac{1}{u}\left(1 - \left(1 - u + o\left(u^2\right)\right)\right)\\ =&\lim_{u\to 0}1+o(u) = 1, \end{align} $$
where we transformed using $u=1/x$, and we used the fact that $\sqrt[3]{1+w} = 1 + \frac{w}{3} + o\left(w^2\right)$.
