1. The problem is to evaluate a limit where direct substitution results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.\n\n2. L'Hopital's theorem states that if $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ or both limits are $\pm \infty$, then:\n$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$\nprovided the latter limit exists.\n\n3. To apply L'Hopital's rule, differentiate numerator and denominator separately.\n\n4. Example: Evaluate $\lim_{x \to 0} \frac{\sin x}{x}$. Direct substitution gives $\frac{0}{0}$.\n\n5. Differentiate numerator and denominator:\n$$f'(x) = \cos x, \quad g'(x) = 1$$\n\n6. Apply L'Hopital's rule:\n$$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1$$\n\n7. Thus, the limit is 1.\n\nThis method helps solve limits that are otherwise indeterminate by simplifying the ratio of derivatives.