Given the function $\displaystyle f(x)=\frac{3x^4+4}{x^3-3x}$ I have found, by long division, that there is a slant asymptote, y=3x.

Further, I wanted to find the same result by the classic limit definition.

$\displaystyle \lim_{x \mapsto \infty}\left [ \frac{3x^4+4}{x^3-3x} \right ]$

$\displaystyle \lim_{x \mapsto \infty}\left [ \frac{x^4(1+\frac{4}{x^4})}{x^4(\frac{1}{x}-\frac{3}{x^3})} \right ]$

$\displaystyle \lim_{x \mapsto \infty}\left [ \frac{1+\frac{4}{x^4}}{\frac{1}{x}-\frac{3}{x^3}} \right ]=\frac{1+0}{0-0}$

I ended up in an indeterminate form.The result should had be the same!What's wrong?