How to find the following limit without using l'Hopital's rule: $\lim_{{x \to 0}} \frac{1 - \cos 3x}{\cos^2 5x - 1}$

Question

\begin{align*} \lim_{{x \to 0}} \frac{1 - \cos 3x}{\cos^2 5x - 1} &= \lim_{{x \to 0}} \frac{1 - \cos 3x}{-\sin^2 5x} \\ &= \lim_{{x \to 0}} (1 - \cos 3x) \cdot \lim_{{x \to 0}} \frac{1}{-\sin^2 5x} \\ &= \lim_{{x \to 0}} (1 - \cos 3x) \cdot (-1) \left( \lim_{{x \to 0}} \frac{5x}{\sin 5x} \right)^2 \cdot \frac{1}{5x}=(0/0) \end{align*}

Answer 1

First Solution: $$\lim_{{x \to 0}} \frac{1 - \cos 3x}{\cos^2 5x - 1} =\lim_{{x \to 0}} \frac{1 - (1-2\sin^2 \frac{3x}{2})}{-(1-\cos^2 5x)} =$$ $$= \lim_{{x \to 0}} \frac{2\sin^2 \frac{3x}{2}}{-\sin^2 5x} =$$ $$= \frac{2\times \frac{3}{2}\times \frac{3}{2}}{(-1)\times 5\times 5} = -\frac{9}{50}$$

Second Solution: $$\lim_{{x \to 0}} \frac{1 - \cos 3x}{\cos^2 5x - 1} =$$ $$=\lim_{{x \to 0}} \frac{(1 - \cos 3x) \times (1 + \cos 3x)}{-(1-\cos^2 5x)\times (1 + \cos 3x)} =$$ $$= \lim_{{x \to 0}} \frac{(1 - \cos^2 3x)}{-\sin^2 5x \times (1 + \cos 3x)} = \lim_{{x \to 0}} \frac{ \sin^2 3x}{(-\sin^2 5x) \times (1 + \cos 3x)} =$$ $$= \frac{3\times 3}{(-1)\times 5\times 5 \times 2} = -\frac{9}{50}$$

Answer 2

By Taylor's formula, $$\cos x= 1-x^2/2+O(x^4)$$ as $x\to 0$. So, $$\frac{1-\cos 3x}{\cos^2 5x-1}=\frac{9x^2/2+O(x^4)}{-25x^2+O(x^4)}$$ as $x\to 0$. The limit is $-9/50$.