Limit with parameter "a" If the function is : 
$$f(x)=\sqrt[3]{x^3-ax^2+1}-\sqrt[3]{x^3-x^2+1}.$$
Determine the parameter "$a$" so that $\lim_{x\rightarrow \infty} f(x)=\dfrac 13$. 
I have tried solving this limit having in mind that "$a$" is some constant and I get $0$ over something, so I calculate that the limit of $f(x)$ is zero. I've started off by multiplying with the conjugate so I get rid of the roots in the numerator, but I am stuck and I don't know how to proceed. Please help...

$$
f(x) = x\sqrt[3]{1-\frac{a}{x} +\frac{1}{x^3}} - x\sqrt[3]{1-\frac{1}{x} +\frac{1}{x^3}} 
$$
expanding the cube radicals we find
$$
f(x) = x\left(1-\frac{1}{3}\left(\frac{a}{x} -\frac{1}{x^3}\right) + O\left(\frac{1}{x^2}\right)\right) - x\left(1-\frac{1}{3}\left(\frac{1}{x} -\frac{1}{x^3}\right) + O\left(\frac{1}{x^2}\right)\right) = -\frac{1}{3}(a-1)+O\left(\frac{1}{x}\right)
$$
thus
$$
\lim_{x\to\infty}f(x) = \lim_{x\to\infty}-\frac{1}{3}(a-1)+O\left(\frac{1}{x}\right) = \frac{1}{3}(1-a)
$$
therefore we require
$$
\frac{1}{3}(1-a) = \frac{1}{3} \implies a = 0
$$
