Bài 5.12 trang 118 SGK Toán 11 tập 1 - Kết nối tri thức

Đề bài

Tính các giới hạn sau:

a) \(\mathop {{\rm{lim}}}\limits_{x \to  + \infty } \frac{{1 - 2x}}{{\sqrt {{x^2} + 1} }}\)

b) \(\mathop {{\rm{lim}}}\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + x + 2}  - x} \right)\)

Quảng cáo

Lời giải chi tiết

Vì \(x \to  + \infty \) nên \(x > 0\), suy ra \(\left| x \right| = x\).

a) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{1 - 2x}}{{\sqrt {{x^2} + 1} }} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {\frac{1}{x} - 2} \right)}}{{\sqrt {{x^2}\left( {1 + \frac{1}{{{x^2}}}} \right)} }} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {\frac{1}{x} - 2} \right)}}{{\left| x \right|\sqrt {\left( {1 + \frac{1}{{{x^2}}}} \right)} }}\)

\( = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {\frac{1}{x} - 2} \right)}}{{x\sqrt {1 + \frac{1}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\frac{1}{x} - 2}}{{\sqrt {1 + \frac{1}{{{x^2}}}} }} = \frac{{0 - 2}}{{\sqrt {1 + 0} }} =  - 2\).

b) \(\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + x + 2}  - x} \right) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\left( {\sqrt {{x^2} + x + 2}  - x} \right)\left( {\sqrt {{x^2} + x + 2}  + x} \right)}}{{\sqrt {{x^2} + x + 2}  + x}}\)

\( = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\left( {{x^2} + x + 2} \right) - {x^2}}}{{\sqrt {{x^2} + x + 2}  + x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x + 2}}{{\sqrt {{x^2} + x + 2}  + x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {1 + \frac{2}{x}} \right)}}{{\sqrt {{x^2}\left( {1 + \frac{1}{x} + \frac{2}{{{x^2}}}} \right)}  + x}}\)

\( = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {1 + \frac{2}{x}} \right)}}{{\left| x \right|\sqrt {1 + \frac{1}{x} + \frac{2}{{{x^2}}}}  + x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {1 + \frac{2}{x}} \right)}}{{x\sqrt {1 + \frac{1}{x} + \frac{2}{{{x^2}}}}  + x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {1 + \frac{2}{x}} \right)}}{{x\left[ {\sqrt {1 + \frac{1}{x} + \frac{2}{{{x^2}}}}  + 1} \right]}}\)

\( = \mathop {\lim }\limits_{x \to  + \infty } \frac{{1 + \frac{2}{x}}}{{\sqrt {1 + \frac{1}{x} + \frac{2}{{{x^2}}}}  + 1}} = \frac{{1 + 0}}{{\sqrt {1 + 0 + 0}  + 1}} = \frac{1}{2}\).