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How to calculate $\lim\limits_{x\to 0} \frac{[\sin{x}-x][\cos({3x})-1]}{x[e^x -1]^4}$ without using L'Hôpital's Rule?

I tried Taylor expansion but I couldn't solve the resulting summations. I also tried expanding them out but there were way too many terms.

What is a valid way to solve the question?

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    $\begingroup$ Taylor works well. The first term on top is about $-\frac{1}{6}x^3$, the second term is about $-\frac{9}{2}x^2$, and the bottom is about $x^5$. $\endgroup$ – André Nicolas Apr 17 '16 at 5:20
  • $\begingroup$ Instead of taking the whole infinite series, take a finite part with error term. $\endgroup$ – DanielWainfleet Apr 17 '16 at 5:40
  • $\begingroup$ See math.stackexchange.com/questions/387333/… $\endgroup$ – lab bhattacharjee Apr 17 '16 at 5:46
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Very rough analysis with Taylor series: $\sin x - x$ vanishes to third order and $\cos(3x) - 1$ vanishes to second order. The denominator vanishes to fifth order, so the limit should be a non-zero real number.

To be much more precise,

$$\sin x - x = -\frac 1 6 x^3 + O(x^5)$$ $$\cos(3x) - 1 = -\frac{(3x)^2}{2} + O(x^4)$$ $$e^x - 1 = x + O(x^2)$$

Convince yourself that all these high order terms can indeed be neglected, and then compute $(-1/6)(-(3/2)^2) = 3/4$.

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  • $\begingroup$ Why $(3/2)^2$ and not $(3)^2/2$? $\endgroup$ – Kamil Jarosz Apr 22 '16 at 4:33

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