I see that you've uploaded an image containing mathematical problems related to limits. Let's work through these questions step by step.

Part A: Multiple Choice

1. $\lim_{x \to 1} \frac{\cos 2x}{\cos x - \sin x}$

We need to evaluate this limit. Using direct substitution: $\lim_{x \to 1} \frac{\cos 2x}{\cos x - \sin x} = \frac{\cos 2}{\cos 1 - \sin 1}.$ Numerically approximating: $\cos 2 \approx -0.4161 \quad \text{and} \quad \cos 1 - \sin 1 \approx 0.2919,$ which gives: $\frac{-0.4161}{0.2919} \approx -1.425.$ Thus, the closest choice is (d) $- \frac{1}{2}$.

2. $\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{\sin x}$

Substituting $x = \frac{\pi}{2}$: $\cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \sin\left(\frac{\pi}{2}\right) = 1,$ so the limit is: $\frac{0^2}{1} = 0.$ The answer is (c) 0.

3. $\lim_{x \to 4} \frac{1 - \tan x}{\sin x - \cos x}$

Substituting $x = 4$: [ \tan(4) \approx 1.1578, \quad \sin(4) \approx -0.7568