1. **State the problem:** Find the limit $$\lim_{x\to 0} \frac{\cos 4x}{\tan x}$$. 2. **Recall important formulas and rules:** - As $x \to 0$, $\cos 4x \to \cos 0 = 1$. - The tangent function near zero behaves like $\tan x \approx x$. - Using the small angle approximation $\tan x \approx x$ helps simplify the limit. 3. **Rewrite the limit using the approximation:** $$\lim_{x\to 0} \frac{\cos 4x}{\tan x} \approx \lim_{x\to 0} \frac{\cos 4x}{x}$$ 4. **Evaluate numerator and denominator separately:** - Numerator: $\cos 4x \to 1$ as $x \to 0$. - Denominator: $x \to 0$. 5. **Direct substitution leads to an indeterminate form $\frac{1}{0}$, so use L'Hôpital's Rule:** - Differentiate numerator: $\frac{d}{dx}(\cos 4x) = -4 \sin 4x$. - Differentiate denominator: $\frac{d}{dx}(\tan x) = \sec^2 x$. 6. **Apply L'Hôpital's Rule:** $$\lim_{x\to 0} \frac{\cos 4x}{\tan x} = \lim_{x\to 0} \frac{-4 \sin 4x}{\sec^2 x}$$ 7. **Evaluate the new limit:** - $\sin 4x \to 0$ as $x \to 0$. - $\sec^2 x = \frac{1}{\cos^2 x} \to 1$ as $x \to 0$. So, $$\lim_{x\to 0} \frac{-4 \sin 4x}{\sec^2 x} = \frac{-4 \cdot 0}{1} = 0$$. **Final answer:** $$\boxed{0}$$