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

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
$$\lim_{x \to \infty} (\frac{3}{x}-1)^{2/3}=((-1)^2)^{-1/3}=1.$$
