1. **Problem:** Find the limit $$\lim_{x \to 1} \frac{\ln x}{x - 1}$$. 2. **Formula and rules:** This is an indeterminate form of type $$\frac{0}{0}$$ as $$\ln 1 = 0$$ and $$1 - 1 = 0$$. We can apply L'Hôpital's Rule which states that if $$\lim_{x \to a} f(x) = 0$$ and $$\lim_{x \to a} g(x) = 0$$, 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 L'Hôpital's Rule:** $$f(x) = \ln x \Rightarrow f'(x) = \frac{1}{x}$$ $$g(x) = x - 1 \Rightarrow g'(x) = 1$$ 4. **Evaluate the new limit:** $$\lim_{x \to 1} \frac{\frac{1}{x}}{1} = \lim_{x \to 1} \frac{1}{x} = 1$$ 5. **Answer:** The limit is 1. --- 1. **Problem:** Find the limit $$\lim_{x \to \frac{\pi}{2}} (\sec x - \tan x)$$. 2. **Recall definitions:** $$\sec x = \frac{1}{\cos x}$$ $$\tan x = \frac{\sin x}{\cos x}$$ 3. **Rewrite the expression:** $$\sec x - \tan x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x}$$ 4. **Evaluate the limit:** As $$x \to \frac{\pi}{2}$$, $$\sin x \to 1$$ and $$\cos x \to 0$$. The numerator $$1 - \sin x \to 0$$ and denominator $$\cos x \to 0$$, so this is an indeterminate form $$\frac{0}{0}$$. 5. **Apply L'Hôpital's Rule:** Differentiate numerator and denominator with respect to $$x$$: $$\frac{d}{dx}(1 - \sin x) = -\cos x$$ $$\frac{d}{dx}(\cos x) = -\sin x$$ 6. **New limit:** $$\lim_{x \to \frac{\pi}{2}} \frac{-\cos x}{-\sin x} = \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\sin x}$$ 7. **Evaluate:** At $$x = \frac{\pi}{2}$$, $$\cos \frac{\pi}{2} = 0$$ and $$\sin \frac{\pi}{2} = 1$$. So the limit is $$\frac{0}{1} = 0$$. 8. **Answer:** The limit is 0. **Final answers:** 1) $$1$$ 2) $$0$$