To prove the limit of $\frac{x^2+2\cos x-2}{x\sin^3 x}$ at zero is $1/12$

Question

To Prove $$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{1}{12}$$ I tried with L'Hospital rule but in vain.

Answer 1

L'Hopital's rule may be used repeatedly until you find an answer:

$$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\lim_{x \to 0}\frac{2x-2\sin x}{\sin^2 x(\sin x+3x\cos x)}$$

$$= \lim_{x \to 0}\frac{2-2\cos x}{ 6\sin ^2(x)\cos (x) - 3x(\sin^3(x)-2\sin(x)\cos^2(x))}$$

$$= \lim_{x \to 0}\frac{2\sin (x)}{ 3(-3\sin^3 (x) + 6\cos^2(x)\sin(x) + x(2\cos^3(x)-7\cos(x)\sin^2(x)))}$$

$$= \lim_{x \to 0}\frac{-2\cos(x)}{ -24\cos^3(x)+84\sin^2(x)\cos(x)+x(60\sin(x)\cos^2(x)-21\sin^3(x))}$$

$$= \frac{-2\cos(0)}{ -24\cos^3(0)+84\sin^2(0)\cos(0)+0(60\sin(0)\cos^2(0)-21\sin^3(0))}$$

$$=\frac{1}{12}$$

Answer 2

Hint. You may use Taylor expansions, as $x \to 0$, $$ \cos x =1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6) $$ $$ \sin x =x+O(x^3) $$ giving $$ x^2+2\cos x-2=\frac{x^4}{12}+O(x^6) $$ $$ x\sin^3 x =x^4+O(x^6) $$ and

$$ \frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{\frac{x^4}{12}+O(x^6)}{x^4+O(x^6)}=\frac{1}{12}+O(x^2) $$

Answer 3

We can use the basic limits \begin{equation*} \lim_{x\rightarrow 0}\frac{\sin x}{x}=1,\ \ \ \ \ \ \ \ \ \lim\limits_{x\rightarrow 0}\dfrac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}}=\dfrac{1% }{24} \end{equation*} as follows \begin{equation*} \lim\limits_{x\rightarrow 0}\frac{x^{2}+2\cos x-2}{x\sin ^{3}x}% =\lim\limits_{x\rightarrow 0}\left( \frac{x}{\sin x}\right) ^{3}\lim\limits_{x\rightarrow 0}\left( \frac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}% }\right) \left( 2\right) =1^{3}\cdot \frac{1}{24}\cdot 2=\frac{1}{12}. \end{equation*}

The basic limits can be computed by repeated use of L'Hospital's rule.

Answer 4

from $$ \cos x = 1 - \frac{x^2}2+\frac{x^4}{24}-\dots $$ we have $$ x^2+2\cos x - 2 = \frac{2x^4}{24}-\dots $$ and $$ \sin x = x-... \\ x\sin^3 x = x^4-... $$

Answer 5

Here is how you can do it via a single application of L'Hospital's Rule. The denominator in the expression can be replaced by $x^4$ via the standard limit $\lim_{x\to 0}(\sin x) /x=1$. And the resulting expression can be rewritten as $$\frac{x^2-4\sin^2(x/2)}{x^4}=\frac{t^2-\sin^2t}{4t^4} =\frac{t-\sin t} {4t^3}\cdot\left(1+\frac{\sin t} {t} \right) $$ using the substitution $x=2t$. The first factor tends to $(1/6)(1/4)=1/24$ via single application of L'Hospital's Rule and the second factor tends to $(1+1)=2$ so that the desired limit is $1/12$.

A little algebraic manipulation combined with standard limits is always a great help when applying the L'Hospital's Rule.