1. **Stating the problem:** Simplify and find the limit of the function $$\frac{1+\sin x - \cos x + \ln(1-x)}{x \tan^2 x}$$ as $x \to 0$ using L'Hôpital's rule. 2. **Recall the formula and rules:** L'Hôpital's rule states that if the limit of $\frac{f(x)}{g(x)}$ as $x \to a$ results in an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, 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. **Check the limit form:** Evaluate numerator and denominator at $x=0$: - Numerator: $1 + \sin 0 - \cos 0 + \ln(1-0) = 1 + 0 - 1 + 0 = 0$ - Denominator: $0 \cdot \tan^2 0 = 0$ So the limit is of the form $\frac{0}{0}$, suitable for L'Hôpital's rule. 4. **Differentiate numerator and denominator:** - Numerator derivative: $$\frac{d}{dx} \left(1 + \sin x - \cos x + \ln(1-x)\right) = \cos x + \sin x - \frac{1}{1-x}$$ - Denominator derivative: $$\frac{d}{dx} \left(x \tan^2 x\right) = \tan^2 x + x \cdot 2 \tan x \cdot \sec^2 x$$ 5. **Evaluate the new limit:** $$\lim_{x \to 0} \frac{\cos x + \sin x - \frac{1}{1-x}}{\tan^2 x + 2x \tan x \sec^2 x}$$ Substitute $x=0$: - Numerator: $\cos 0 + \sin 0 - \frac{1}{1-0} = 1 + 0 - 1 = 0$ - Denominator: $\tan^2 0 + 2 \cdot 0 \cdot \tan 0 \cdot \sec^2 0 = 0 + 0 = 0$ Again, indeterminate $\frac{0}{0}$ form. 6. **Apply L'Hôpital's rule again:** Differentiate numerator and denominator once more. - Numerator second derivative: $$-\sin x + \cos x - \frac{d}{dx} \left(\frac{1}{1-x}\right) = -\sin x + \cos x - \frac{1}{(1-x)^2}$$ - Denominator second derivative: Differentiate $\tan^2 x + 2x \tan x \sec^2 x$: $$2 \tan x \sec^2 x + 2 \tan x \sec^2 x + 2x \left(\sec^2 x \sec^2 x + \tan x \cdot 2 \sec^2 x \tan x\right)$$ Simplify stepwise: $$4 \tan x \sec^2 x + 2x \left(\sec^4 x + 2 \tan^2 x \sec^2 x\right)$$ 7. **Evaluate the second derivative limit at $x=0$:** - Numerator: $$-\sin 0 + \cos 0 - \frac{1}{(1-0)^2} = 0 + 1 - 1 = 0$$ - Denominator: $$4 \tan 0 \sec^2 0 + 2 \cdot 0 \cdot (\sec^4 0 + 2 \tan^2 0 \sec^2 0) = 0 + 0 = 0$$ Still indeterminate. 8. **Apply L'Hôpital's rule a third time:** Differentiate numerator and denominator again. - Numerator third derivative: $$-\cos x - \sin x - \frac{d}{dx} \left(\frac{1}{(1-x)^2}\right) = -\cos x - \sin x - \left(-2 \frac{1}{(1-x)^3}\right) = -\cos x - \sin x + \frac{2}{(1-x)^3}$$ - Denominator third derivative: Differentiate: $$4 \tan x \sec^2 x + 2x (\sec^4 x + 2 \tan^2 x \sec^2 x)$$ Derivative: $$4 (\sec^2 x \sec^2 x + \tan x \cdot 2 \sec^2 x \tan x) + 2 (\sec^4 x + 2 \tan^2 x \sec^2 x) + 2x \frac{d}{dx} (\sec^4 x + 2 \tan^2 x \sec^2 x)$$ At $x=0$, terms with $x$ vanish, so evaluate only the first two terms: $$4 (\sec^4 0 + 2 \tan^2 0 \sec^2 0) + 2 (\sec^4 0 + 2 \tan^2 0 \sec^2 0) = 4 (1 + 0) + 2 (1 + 0) = 4 + 2 = 6$$ 9. **Evaluate the third derivative limit at $x=0$:** - Numerator: $$-\cos 0 - \sin 0 + \frac{2}{(1-0)^3} = -1 - 0 + 2 = 1$$ - Denominator: $$6$$ 10. **Final limit:** $$\lim_{x \to 0} \frac{1+\sin x - \cos x + \ln(1-x)}{x \tan^2 x} = \frac{1}{6}$$ **Answer:** $\boxed{\frac{1}{6}}$