Limit of a rational function Calculate the limit
$$ \lim_{x \to 0} \frac{3x^{2} - \frac{x^{4}}{6}}{(4x^{2} - 8x^{3} + \frac{64x^{4}}{3} )}$$
I divided by the highest degree of x, which is $x^{4}$, further it gave 
$$ \frac{-\frac{1}{6}}{\frac{64}{3}} = \frac{-1}{128}$$ 
which is wrong... what is my error? 
 A: $$ \lim_{x \to 0} \frac{3x^{2} - \dfrac{x^{4}}{6}}{(4x^{2} - 8x^{3} + \dfrac{64x^4}{3} )}$$
$$=\lim_{x\to0}\frac{x^2\left(3-\dfrac{x^2}6\right)}{x^2\left(4-8x+\dfrac{64}3x^2\right)}$$
Cancel out $x^2$ as $x\ne0$ as $x\to0$
Then set $x=0$ as it is no longer of the form $\dfrac00$
A: If your were taking the limit of your function as $x\to \infty$, then your approach would have worked. When $x\to \infty$, we divide numerator and denominator by the highest degree in the denominator. 
However, here you are evaluating a limit as $x\to 0$.  When we have a limit $\lim_{x\to 0} \frac{p(x)}{q{x}}$, as is the case here, we divide numerator and denominator by the lowest degree.
Make that change, and you'll find the correct limit to be $\dfrac 34$.
A: At $0$ we have
$$x^4=o(x^2)\quad\text{and}\quad x^3=o(x^2)$$
so
$$ \lim_{x \to 0} \frac{3x^{2} - \frac{x^{4}}{6}}{(4x^{2} - 8x^{3} + \frac{64x^{4}}{3} )}=\lim_{x \to 0}\frac{3x^2}{4x^2}=\frac34$$
A: You'll see $$\frac{\color{red}{x^2}(3-(x^2/6))}{\color{red}{x^2}(4-8x+(64x^2/3))}=\frac{3-(x^2/6)}{4-8x+(64x^2/3)}\to\frac{3}{4}\ (x\to 0).$$
