1. Find \( \lim_{x \to 7^-} \frac{x^2 + 5x - 19}{x + 7} \). Step 1: Substitute \( x = 7 \) into the expression. $$\frac{7^2 + 5\cdot7 - 19}{7 + 7} = \frac{49 + 35 - 19}{14} = \frac{65}{14}$$ Since the denominator is not zero, the limit is \( \frac{65}{14} \). 2. Find \( \lim_{x \to 7} \frac{x}{2 - \sqrt{4 + x}} \). Step 1: Substitute \( x = 7 \). Denominator: \( 2 - \sqrt{4 + 7} = 2 - \sqrt{11} \neq 0 \). Numerator: \( 7 \). Step 2: Calculate the limit directly. $$\frac{7}{2 - \sqrt{11}}$$ To rationalize the denominator: $$\frac{7}{2 - \sqrt{11}} \cdot \frac{2 + \sqrt{11}}{2 + \sqrt{11}} = \frac{7(2 + \sqrt{11})}{(2)^2 - (\sqrt{11})^2} = \frac{7(2 + \sqrt{11})}{4 - 11} = \frac{7(2 + \sqrt{11})}{-7} = -(2 + \sqrt{11})$$ Therefore, the limit is \( -(2 + \sqrt{11}) \). 3. Find \( \lim_{x \to 7} ( \sqrt{x^2 - 2x + 1} - (x + 3) ) \). Step 1: Simplify inside the square root: $$x^2 - 2x + 1 = (x - 1)^2$$ Step 2: So the expression becomes: $$\sqrt{(x - 1)^2} - (x + 3) = |x - 1| - (x + 3)$$ Step 3: Since \(x \to 7\), and \(7 - 1 = 6 > 0\), so \(|x - 1| = x - 1\). Step 4: Substitute: $$ (x - 1) - (x + 3) = x - 1 - x - 3 = -4 $$ Therefore, the limit is \( -4 \). 4. Find \( \lim_{x \to 0} \frac{\cos 4x - \cos 2x}{x^2} \). Step 1: Use the cosine difference identity: $$\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$$ Step 2: Apply it with \( A = 4x \), \( B = 2x \): $$\cos 4x - \cos 2x = -2 \sin 3x \sin x$$ Step 3: The expression becomes: $$\frac{-2 \sin 3x \sin x}{x^2}$$ Step 4: Rewrite as: $$-2 \left( \frac{\sin 3x}{x} \right) \left( \frac{\sin x}{x} \right)$$ Step 5: Use the limit \( \lim_{x \to 0} \frac{\sin kx}{x} = k \): $$\lim_{x \to 0} \frac{\sin 3x}{x} = 3$$ $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$ Step 6: So the limit is: $$-2 \times 3 \times 1 = -6$$ 5. Find \( \lim_{x \to \pi/4} \frac{\sin 4x - \sin \pi}{\cos 2x} \). Step 1: Note \( \sin \pi = 0 \), so numerator is \( \sin 4x \). Step 2: Substitute \( x = \pi/4 \): $$\sin 4 \cdot \frac{\pi}{4} = \sin \pi = 0$$ $$\cos 2 \cdot \frac{\pi}{4} = \cos \frac{\pi}{2} = 0$$ Step 3: The expression becomes \( \frac{0}{0} \), an indeterminate form. Step 4: Apply L'Hôpital's Rule: Differentiate numerator: $$\frac{d}{dx} \sin 4x = 4 \cos 4x$$ Differentiate denominator: $$\frac{d}{dx} \cos 2x = -2 \sin 2x$$ Step 5: Compute limit of ratio of derivatives at \( x = \pi/4 \): $$\lim_{x \to \pi/4} \frac{4 \cos 4x}{-2 \sin 2x} = \frac{4 \cos \pi}{-2 \sin \frac{\pi}{2}} = \frac{4(-1)}{-2(1)} = \frac{-4}{-2} = 2$$ Therefore, the limit is \( 2 \).