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### 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

Solve limit problems involving trigonometric and algebraic expressions step by step. Learn how to apply limit laws, trigonometric identities, and L'Hopital's Rule to evaluate these limits. Suitable for students in Grades 11-12.

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