# 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?

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?

• 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 $x^5$. – André Nicolas Apr 17 '16 at 5:20
• Instead of taking the whole infinite series, take a finite part with error term. – DanielWainfleet Apr 17 '16 at 5:40
• – lab bhattacharjee Apr 17 '16 at 5:46

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.
$$\sin x - x = -\frac 1 6 x^3 + O(x^5)$$ $$\cos(3x) - 1 = -\frac{(3x)^2}{2} + O(x^4)$$ $$e^x - 1 = x + O(x^2)$$
Convince yourself that all these high order terms can indeed be neglected, and then compute $(-1/6)(-(3/2)^2) = 3/4$.
• Why $(3/2)^2$ and not $(3)^2/2$? – Kamil Jarosz Apr 22 '16 at 4:33