Evaluating a Limit (without L'Hopital's rule)

Question

I just got this limit problem. It is easy but there is an instruction to solve this limit without L'Hopital's rule. How can I derived this?

$$\lim_{x\to 0} \left(\frac{1}{\sin^2(x)} + \frac{\cos^2(x)}{\sin^2(x)} - \frac{2}{x^2}\right)$$

Answer 1

$\begin{align} \lim_{x\to 0} \left(\frac{1}{\sin^2(x)} + \frac{\cos^2(x)}{\sin^2(x)} - \frac{2}{x^2}\right)&=\lim_{x\to 0} \left(\frac {x^2+x^2\cos^2x-2\sin^2x}{x^2\sin^2x} \right)\\ &=\lim_{x\to 0} \left(\frac {2x^2-x^2\sin^2x-2\sin^2x}{x^2\sin^2x} \right)\\ &=-1+\lim_{x\to 0} \left(\frac {2(x-\sin x)(x+\sin x)}{x^2\sin^2x} \right)\\ &=-1+\lim_{x\to 0} \left(\frac {2(x-(x-\frac{x^3}6+O(x^5)))(x+x+O(x^3))}{x^2(x+O(x^3))^2} \right)\\ &=-1+\lim_{x\to 0} \left(\frac{(\frac{x^3}3+O(x^5))(2x+O(x^3))}{x^4+O(x^6)} \right)\\ &=-1+\lim_{x\to 0} \left(\frac{\frac23x^4+O(x^6)}{x^4+O(x^6)} \right)\\ &=-1+\lim_{x\to 0} \left(\frac{\frac23+O(x^2)}{1+O(x^2)} \right)\\ &=-1+\frac23\\ &=-\frac13 \end{align}$

Answer 2

\begin{eqnarray} \mathcal L &=& \lim_{x\to 0}\left(\frac1{\sin^2x} + \frac{\cos^2x}{\sin^2x}-\frac2{x^2}\right)=\\ &=&\lim_{x\to 0} \left(\frac2{\sin^2x}-1-\frac2{x^2}\right)=\\ &=& 2\lim_{x\to 0}\left(\frac{x^2-\sin^2x}{x^2\sin^2x}\right)-1=\\ &=&2\lim_{x\to 0}\left(\frac{x^2-\sin^2x}{x^4}\right)-1=\\ &=&2\lim_{x\to 0}\left(\frac{x+\sin x}{x}\cdot\frac{x-\sin x}{x^3}\right)-1=\\ &=& 4\underbrace{\lim_{x\to 0}\left(\frac{x-\sin x}{x^3}\right)}_{\mathcal{L_1}}-1. \end{eqnarray} There are nice ways to find $$\mathcal L_1 = \frac16,$$ such as this one. Hence $\mathcal L = -\frac13$.