Limit without hospital's rule $\lim_{x\rightarrow 0} (\frac{1}{x^2} - \frac{x}{\sin^3(x)})$

Question

Calcule $\lim_{x\rightarrow 0} (\frac{1}{x^2} - \frac{x}{\sin^3(x)})$ without using L'Hospital

I first was trying to replace $x = \frac{\pi}{2} + u$ but I can't find a rule or eliminate anything. After that I try using cubes:

$\frac{\sin^3(x) - x^3}{x^2\sin^3(x)} = \frac{(\sin(x)-x)(\sin^2(x)+x^2+x\sin(x))}{x^2\sin^3(x)}$ but again i couldn't find any rule or formula.

How can I solve this problem without using L'Hospital? Thank you very much in advance

Answer 1

$$ \begin{align} \lim_{x\to 0}\left(\frac{1}{x^2}-\frac{x}{\sin^3x}\right) &= \lim_{x\to 0}\left(\frac{\sin^3x-x^3}{x^2\sin^2x}\right) \\ &= \lim_{x\to 0}\left(\frac{x^3}{\sin^3x}\right)\left(\frac 1{x^2}\right)\left(\frac{\sin^3x}{x^3}-1\right) \\ &=1\cdot\lim_{x\to 0}\left(\frac{1}{x^2}\right)\left(\frac{\sin x}{x}-1\right)\left(\frac{\sin^2x}{x^2}+\frac{\sin x}{x}+1\right) \\ &=1\cdot\lim_{x\to 0}\left(\frac{\sin x-x}{x^3}\right)\cdot3 \\ &=1\cdot\left(\frac {-1}6\right)\cdot3 \\ &=\frac{-1}{2} \end{align} $$

Answer 2

We know $\lim_{x\to0}\frac{\sin x}x=1$, so your limit is $$ \lim_{x\to0}\frac{1+\left(\frac{\sin x}{x}\right)^{-3}}{x^2} $$

And it's numerator goes to 1, and denominator goes to 0.

So your limit is $\infty$(especially, positive valued.).