1. Problem: Calculate \(\lim_{\theta \to 0} \frac{\tan 5\theta}{\sin 2\theta}\). Step 1: Using the standard limit \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) and \(\lim_{x \to 0} \frac{\tan x}{x} = 1\), rewrite the expression: $$\lim_{\theta \to 0} \frac{\tan 5\theta}{\sin 2\theta} = \lim_{\theta \to 0} \frac{\tan 5\theta}{5\theta} \times \frac{5\theta}{2\theta} \times \frac{2\theta}{\sin 2\theta}$$ Step 2: Simplify each part using limits: - \(\lim_{\theta \to 0} \frac{\tan 5\theta}{5\theta} = 1\) - \(\frac{5\theta}{2\theta} = \frac{5}{2}\) - \(\lim_{\theta \to 0} \frac{2\theta}{\sin 2\theta} = 1\) Step 3: Multiply results: $$1 \times \frac{5}{2} \times 1 = \frac{5}{2}$$ Answer: $$\boxed{\frac{5}{2}}$$ 2. Problem: Calculate \(\lim_{t \to 0} \frac{1 - \cos 2t}{t^2}\). Step 1: Recall the identity \(1 - \cos 2t = 2\sin^2 t\). Step 2: Substitute into the limit: $$\lim_{t \to 0} \frac{2 \sin^2 t}{t^2} = 2 \times \lim_{t \to 0} \left( \frac{\sin t}{t} \right)^2$$ Step 3: Since \(\lim_{t \to 0} \frac{\sin t}{t} = 1\), then $$2 \times 1^2 = 2$$ Answer: $$\boxed{2}$$ 3. Problem: Calculate \(\lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{1 - \tan x}\). Step 1: Plug \(x = \frac{\pi}{4}\) directly: \(\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}\), \(\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}\), \(\tan \frac{\pi}{4} = 1\) Step 2: Numerator: $$\sin x - \cos x = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0$$ Denominator: $$1 - \tan x = 1 - 1 = 0$$ Step 3: Since direct substitution yields \(\frac{0}{0}\), use L'Hôpital's Rule: Differentiate numerator: $$\frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x$$ Differentiate denominator: $$\frac{d}{dx}(1 - \tan x) = -\sec^2 x$$ Step 4: Evaluate derivatives at \(x=\frac{\pi}{4}\): $$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \sec^2 \frac{\pi}{4} = \left( \frac{1}{\cos \frac{\pi}{4}} \right)^2 = 2$$ Step 5: Substitute: $$\lim_{x \to \frac{\pi}{4}} \frac{\cos x + \sin x}{-\sec^2 x} = \frac{ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} }{-2} = \frac{\sqrt{2}}{-2} = - \frac{\sqrt{2}}{2}$$ Answer: $$\boxed{- \frac{\sqrt{2}}{2}}$$ 4. Problem: Calculate \(\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{1 - \sin x}\). Step 1: Use the identity for the limit near \(x = \frac{\pi}{2}\): The numerator at \(x=\frac{\pi}{2}\) is \(\cos^2 \frac{\pi}{2} = 0\). Denominator at \(x=\frac{\pi}{2}\) is \(1 - \sin \frac{\pi}{2} = 1 - 1 = 0\). Step 2: Apply L'Hôpital's Rule. Differentiate numerator: $$\frac{d}{dx} \cos^2 x = 2 \cos x (-\sin x) = -2 \sin x \cos x$$ Differentiate denominator: $$\frac{d}{dx} (1 - \sin x) = -\cos x$$ Step 3: Evaluate derivatives at \(x = \frac{\pi}{2}\): \(\sin \frac{\pi}{2} = 1\), \(\cos \frac{\pi}{2} = 0\) Step 4: Substitute: $$\frac{-2 (1)(0)}{-0} = \frac{0}{0}$$ Step 5: Apply L'Hôpital's Rule again. Differentiate numerator again: $$\frac{d}{dx} (-2 \sin x \cos x) = -2(\cos^2 x - \sin^2 x)$$ Differentiate denominator again: $$\frac{d}{dx} (-\cos x) = \sin x$$ Step 6: Evaluate at \(x=\frac{\pi}{2}\): \(\cos \frac{\pi}{2} = 0\), \(\sin \frac{\pi}{2} = 1\) Numerator: $$-2 (0 - 1) = -2 (-1) = 2$$ Denominator: $$1$$ Step 7: Compute limit: $$\frac{2}{1} = 2$$ Answer: $$\boxed{2}$$