How to find $\lim\limits_{x\to0}\frac{x^2+2\cos x-2}{x \sin^3x}$

Question

How to evaluate $$\lim_{x\to0}\frac{x^2+2\cos x-2}{x \sin^3x}?$$

I tried using L'Hospital but it seemed that there is an infinite loop of 0/0 form

Answer 1

Using that $\cos x= 1-\frac{x^2}2+\frac{x^4}{24}+o(x^6)$, and $\sin x=x+o(x^3)$, $$ \frac{(x^2+2\cos x-2)}{(x \sin^3x)} =\frac{x^4/12+o(x^6)}{x(x+o(x^3))^3} =\frac{x^4/12+o(x^6)}{x^4+o(x^6)} =\frac{1/12+o(x^2)}{1+o(x^2)}\xrightarrow[x\to0]{}\frac1{12}. $$

Answer 2

Use Taylor series for $\cos x$ about $x=0$

$$\cos x = 1 -x^2/2+x^4/24 -... $$

Thus $$\lim_{x\to0}\frac{(x^2+2\cos x-2)}{(x \sin^3x)}=\lim_{x\to0}\frac{x^4}{12x \sin^3x} =\frac {1}{12}$$

Answer 3

All that is needed is $\sin(x) =x-x^3/6+O(x^5) =x(1-x^2/6+O(x^4)) $ so $\sin^2(x) =x^2(1-x^2/3+O(x^4)) $ and $\sin^3(x) =x^3(1-x^2/2+O(x^4)) $.

$\begin{array}\\ \dfrac{(x^2+2\cos x-2)}{x \sin^3x} &=\dfrac{x^2+2(\cos x-1)}{x \sin^3x}\\ &=\dfrac{x^2+2(-2\sin^2(x/2)}{x \sin^3x}\\ &=\dfrac{x^2-4\sin^2(x/2)}{x \sin^3x}\\ &=\dfrac{x^2-4(x/2)^2(1-(x/2)^2/3+O(x^4))}{x\cdot x^3(1-x^2/2+O(x^4))}\\ &=\dfrac{x^2-x^2(1-x^2/12+O(x^4))}{x^4(1-x^2/2+O(x^4))}\\ &=\dfrac{x^4/12+O(x^6))}{x^4(1-x^2/2+O(x^4))}\\ &=\dfrac{1/12+O(x^2))}{1-x^2/2+O(x^4)}\\ &\to 1/12\\ \end{array} $

Answer 4

For the L'Hospital's rule, use $\sin x\sim x \Rightarrow \sin^3x \sim x^3, x\to0$: $$\begin{align}\lim_{x\to0}\frac{x^2+2\cos x-2}{x \sin^3x}&=\lim_{x\to0}\frac{x^2+2\cos x-2}{x^4}\stackrel{LR}=\\ &=\lim_{x\to0}\frac{2x-2\sin x}{4x^3}\stackrel{LR}=\\ &=\lim_{x\to0}\frac{2-2\cos x}{12x^2}\stackrel{LR}=\\ &=\lim_{x\to0}\frac{2\sin x}{24x}=\\ &=\frac{1}{12}.\end{align}$$