evaluate $\lim _{x\to \infty} (3x^2-x^3)^{\frac{1}{3}}+x$ 
$$\lim _{x\to \infty} (3x^2-x^3)^{\frac{1}{3}}+x$$

can I look at $\lim\limits_{x\to \infty} (3^{\frac{1}{3}}x^{\frac{2}{3}}-x+x)$?
 A: HINT:
$$\left(1+z\right)^{1/3}=1+\frac13 z+O(z^2) \tag 1$$
SPOILER ALERT:  Scroll over the highlighted area to reveal the full solution

 $$\begin{align}x+\left(3x^2-x^3\right)^{1/3}&=x-x\left(1-\frac{3x^2}{x^3}\right)^{1/3}\\\\&=x-x\left(1-\frac3x\right)^{1/3}\\\\&=x-x\left(1-\frac1x+O\left(\frac1{x^2}\right)\right)\\\\&=1+O\left(\frac1x\right)\to 1\,\,\text{as}\,\,x\to \infty\end{align}$$

A: Set $x=\frac1h$, then the limit transforms to
$$
\lim_{h\to0}\frac{1-(1-3h)^{1/3}}h
$$
which is the derivative of $f(x)=(1+3x)^{1/3}$ at $x=0$. 
Or 3 times the derivative of $f(x)=x^{1/3}$ at $x=1$
A: You can, but it'll probably give you the wrong answer. In general, it's not true that 
$$
(a + b)^k = a^k + b^k
$$
but you're using that idea in trying to simplify the parenthesized expression. 
To see that it doesn't generally work, look at 
$$
(2 + 2)^3 = 64 \\
2^3 + 2^3 = 16. 
$$
A: By factoring, the expression becomes $(x^2(3-x))^{1/3}+x=x^{2/3}(3-x)^{1/3}+x$. Now we take a seemingly random detour and multiply the numerator and denominator by $1/x$, giving $$\lim_{x \to \infty}\frac{x^{-1/3}(3-x)^{1/3}+1}{1/x}= \lim_{x \to \infty}\frac{x^{(-1){1/3}}(3-x)^{1/3}+1}{1/x}=\lim_{x \to \infty}\frac{(\frac{3-x}{x})^{1/3}+1}{1/x}$$
Taking x $\rightarrow$ $\infty$ gives the indeterminate form $\frac{0}{0}$.
Applying L'Hopital's Rule gives
$$\lim_{x \to \infty}\frac{\frac{1}{3}(\frac{3-x}{x})^{-2/3}(\frac{-3}{x^2})}{-1/x^2}.$$
After cancelling like terms, we find
$$\lim_{x \to \infty} (\frac{3}{x}-1)^{2/3}=((-1)^2)^{-1/3}=1.$$
