Limit of $\frac{1}{x^2}-\frac{1}{\sin^2(x)}$ as $x$ approaches $0$ [duplicate]

Solution 1:

$$\begin{aligned}L &= \lim_{x \to 0}\frac{1}{x^{2}} - \frac{1}{\sin^{2}x}\\ &= \lim_{x \to 0}\frac{\sin^{2} x - x^{2}}{x^{2}\sin^{2}x}\\ &= \lim_{x \to 0}\frac{(\sin x - x)(\sin x + x)}{x^{3}\cdot x}\cdot\frac{x^{2}}{\sin^{2}x}\\ &= \lim_{x \to 0}\frac{(\sin x - x)(\sin x + x)}{x^{3}\cdot x}\\ &= \lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\lim_{x \to 0}\frac{\sin x + x}{x}\\ &= \lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\left(\lim_{x \to 0}\frac{\sin x}{x} + 1\right)\\ &= 2\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\\ &= 2\lim_{x \to 0}\frac{\cos x - 1}{3x^{2}}\text{ (applying L'Hospital's Rule)}\\ &= -\frac{2}{3}\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot\frac{1 + \cos x}{1 + \cos x}\\ &= -\frac{2}{3}\lim_{x \to 0}\frac{1 - \cos^{2} x}{x^{2}(1 + \cos x)}\\ &= -\frac{2}{3}\lim_{x \to 0}\frac{\sin^{2} x}{x^{2}}\cdot\frac{1}{1 + \cos x}\\ &= -\frac{2}{3}\cdot 1\cdot\frac{1}{2} = -\frac{1}{3}\end{aligned}$$

Thus as mentioned by OP in comments, it is doable without Taylor series. In fact the limit of $(\sin x - x)/x^{3}$ is also doable without L'Hospital Rule, but it requires more work as shown by user robjohn in a beautiful answer.