	Minh Hưng \| Chat Online
	18/12/2025 13:18:09
	Toán học - Lớp 11 \| Toán học \| Lớp 11

2. \(\lim_{x \to \frac{1}{2}} \frac{2x^2 - 5x + 2}{1 - 2x}\) 5. \(\lim_{x \to -3} \frac{x + 3}{x^2 - 9}\) 8. \(\lim_{x \to -2} \frac{x^3 + 3x^2 - 9x - 2}{-x^3 + x + 6}\) 11. \(\lim_{x \to 0} \frac{\sqrt{1 + 2x} - 1}{2x}\) 14. \(\lim_{x \to 0} \frac{\sqrt{1 + x^2} - 1}{\sqrt{x^2 + 16} - 4}\) 17. \(\lim_{x \to -1} \frac{2x + \sqrt{3x^2} + 1}{x^2 - 1}\) 20. \(\lim_{x \to 4} \frac{\sqrt{2x + 9} - x - 5}{\sqrt[3]{x + 5} + \sqrt[3]{x + 3}}\)

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tính các giới hạn sau
2. \(\lim_{x \to \frac{1}{2}} \frac{2x^2 - 5x + 2}{1 - 2x}\)
5. \(\lim_{x \to -3} \frac{x + 3}{x^2 - 9}\)
8. \(\lim_{x \to -2} \frac{x^3 + 3x^2 - 9x - 2}{-x^3 + x + 6}\)
11. \(\lim_{x \to 0} \frac{\sqrt{1 + 2x} - 1}{2x}\)
14. \(\lim_{x \to 0} \frac{\sqrt{1 + x^2} - 1}{\sqrt{x^2 + 16} - 4}\)
17. \(\lim_{x \to -1} \frac{2x + \sqrt{3x^2} + 1}{x^2 - 1}\)
20. \(\lim_{x \to 4} \frac{\sqrt{2x + 9} - x - 5}{\sqrt[3]{x + 5} + \sqrt[3]{x + 3}}\)