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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?

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1  
Divide top and bottom by $x^2$. –  André Nicolas Aug 25 at 18:01
2  
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 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 at 18:06

4 Answers 4

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 $\dfrac 34$.

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

$$=\lim_{x\to0}\frac{x^2\left(3-\dfrac{x^2}6\right)}{x^2\left(4-8x+\dfrac{64}3x^2\right)}$$

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

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

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

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

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