1. The problem asks to find the value of $k$ for which the function $$f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ k, & x = 0 \end{cases}$$ is continuous at $x=0$. 2. To ensure continuity at $x=0$, the limit of $f(x)$ as $x \to 0^-$ must equal $f(0)=k$. 3. Calculate the left-hand limit: $$\lim_{x \to 0^-} \frac{1-\cos 4x}{x^2}$$ 4. Use the Taylor series expansion for cosine near zero: $$\cos 4x = 1 - \frac{(4x)^2}{2} + O(x^4) = 1 - 8x^2 + O(x^4)$$ 5. Substitute this into the expression: $$\frac{1 - (1 - 8x^2 + O(x^4))}{x^2} = \frac{8x^2 + O(x^4)}{x^2} = 8 + O(x^2)$$ 6. Taking the limit $x \to 0$, the limit is $8$. 7. Therefore, $k = 8$ for continuity at $x=0$. --- 8. Next problem: The function $$f(x) = \begin{cases} 2x + 2, & x < 2 \\ k, & x = 2 \\ 3x, & x > 2 \end{cases}$$ must be continuous at $x=2$. 9. For continuity at $x=2$, the left and right limits as $x \to 2$ must be equal to $f(2) = k$. 10. Compute left-hand limit: $$\lim_{x \to 2^-} 2x + 2 = 2(2) + 2 = 6$$ 11. Compute right-hand limit: $$\lim_{x \to 2^+} 3x = 3(2) = 6$$ 12. Since both limits equal 6, set $k = 6$ to ensure continuity. --- 13. Last problem: The function $$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases}$$ must be continuous at $x = \frac{\pi}{2}$. 14. Continuity requires $$\lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = 3$$ 15. Evaluate the limit: $$\lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x}$$ 16. Both numerator and denominator approach zero: $$\cos \frac{\pi}{2} = 0, \quad \pi - 2 \cdot \frac{\pi}{2} = \pi - \pi = 0$$ 17. Apply L'Hôpital's Rule: $$\lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x} = \lim_{x \to \frac{\pi}{2}} \frac{-k \sin x}{-2} = \lim_{x \to \frac{\pi}{2}} \frac{k \sin x}{2}$$ 18. Substitute $x = \frac{\pi}{2}$: $$\frac{k \cdot 1}{2} = \frac{k}{2}$$ 19. Set equal to 3 for continuity: $$\frac{k}{2} = 3 \implies k = 6$$ --- Final answers: - For the first function, $k = 8$. - For the second, $k=6$. - For the third, $k=6$.