Bài 6d (SGK-T133):Tính: \[\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 1} + x}}{{5 – 2x}}\]

Giải: $\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 1} + x}}{{5 – 2x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {\sqrt {1 + \frac{1}{{{x^2}}}} + 1} \right)}}{{5 – 2x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {\sqrt {1 + \frac{1}{{{x^2}}}} + 1} \right)}}{{\frac{5}{x} Đọc tiếp…

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