Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Divide top and bottom by $x^2$. – André Nicolas Aug 25 '14 at 18:01
I just realised that when looking at the question... I should use the highest degree when $ x \to \inf$ and the lowest when we have to zero. :/ – Ara Aug 25 '14 at 18:02
If you divide by top and bottom by $x^4$, most of the terms blow up as $x$ approaches $0$. So it is not clear what the limit is, if any. – André Nicolas Aug 25 '14 at 18:06

$$ \lim_{x \to 0} \frac{3x^{2} - \dfrac{x^{4}}{6}}{(4x^{2} - 8x^{3} + \dfrac{64x^4}{3} )}$$


Cancel out $x^2$ as $x\ne0$ as $x\to0$

Then set $x=0$ as it is no longer of the form $\dfrac00$

share|cite|improve this answer

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 $\dfrac 34$.

share|cite|improve this answer

At $0$ we have $$x^4=o(x^2)\quad\text{and}\quad x^3=o(x^2)$$ so $$ \lim_{x \to 0} \frac{3x^{2} - \frac{x^{4}}{6}}{(4x^{2} - 8x^{3} + \frac{64x^{4}}{3} )}=\lim_{x \to 0}\frac{3x^2}{4x^2}=\frac34$$

share|cite|improve this answer

You'll see $$\frac{\color{red}{x^2}(3-(x^2/6))}{\color{red}{x^2}(4-8x+(64x^2/3))}=\frac{3-(x^2/6)}{4-8x+(64x^2/3)}\to\frac{3}{4}\ (x\to 0).$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.