1. **Problem:** Find the limit $$\lim_{x \to 0} \frac{\cosh x - \cos x}{x^2}$$ using L'Hopital-Bernoulli rule. 2. **Recall:** L'Hopital's rule states that if $$\lim_{x \to a} f(x) = 0$$ and $$\lim_{x \to a} g(x) = 0$$ or both limits are infinite, then $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ provided the latter limit exists. 3. **Apply:** Here, numerator and denominator both approach 0 as $$x \to 0$$. 4. Differentiate numerator and denominator: $$f(x) = \cosh x - \cos x \implies f'(x) = \sinh x + \sin x$$ $$g(x) = x^2 \implies g'(x) = 2x$$ 5. New limit: $$\lim_{x \to 0} \frac{\sinh x + \sin x}{2x}$$ 6. Again numerator and denominator approach 0, apply L'Hopital's rule again: $$f'(x) = \cosh x + \cos x$$ $$g'(x) = 2$$ 7. Evaluate limit: $$\lim_{x \to 0} \frac{\cosh x + \cos x}{2} = \frac{\cosh 0 + \cos 0}{2} = \frac{1 + 1}{2} = 1$$ **Final answer:** $$1$$