Câu 19 trang 226 SGK Đại số và Giải tích 11 Nâng cao

Tính giới hạn của các hàm số sau :

a. \(\mathop {\lim }\limits_{x \to  - 1} {{{x^2} + x + 10} \over {{x^3} + 6}}\)

b. \(\mathop {\lim }\limits_{x \to  - 5} {{{x^2} + 11x + 30} \over {25 - {x^2}}}\)

c. \(\mathop {\lim }\limits_{x \to  - \infty } {{{x^6} + 4{x^2} + x - 2} \over {{{\left( {{x^3} + 2} \right)}^2}}}\)

d. \(\mathop {\lim }\limits_{x \to  + \infty } {{{x^2} + x - 40} \over {2{x^5} + 7{x^4} + 21}}\)

e. \(\mathop {\lim }\limits_{x \to  - \infty } {{\sqrt {2{x^4} + 4{x^2} + 3} } \over {2x + 1}}\)

f. \(\mathop {\lim }\limits_{x \to  + \infty } \left( {2x + 1} \right)\sqrt {{{x + 1} \over {2{x^3} + x}}} \)

g. \(\mathop {\lim }\limits_{x \to  + \infty } \sqrt {9{x^2} + 11x - 100} \)

h. \(\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {5{x^2} + 1}  - x\sqrt 5 } \right)\)

i. \(\mathop {\lim }\limits_{x \to  + \infty } {1 \over {\sqrt {{x^2} + x + 1}  - x}}\)

a. \(\mathop {\lim }\limits_{x \to  - 1} {{{x^2} + x + 10} \over {{x^3} + 6}} = {{1 + \left( { - 1} \right) + 10} \over { - 1 + 6}} = 2\)

b. \(\mathop {\lim }\limits_{x \to  - 5} {{{x^2} + 11x + 30} \over {25 - {x^2}}} = \mathop {\lim }\limits_{x \to  - 5} {{\left( {x + 5} \right)\left( {x + 6} \right)} \over {\left( {5 - x} \right)\left( {5 + x} \right)}} = \mathop {\lim }\limits_{x \to  - 5} {{x + 6} \over {5 - x}} = {1 \over {10}}\)

c. \(\mathop {\lim }\limits_{x \to  - \infty } {{{x^6} + 4{x^2} + x - 2} \over {{{\left( {{x^3} + 2} \right)}^2}}} = \mathop {\lim }\limits_{x \to  - \infty } {{1 + {4 \over {{x^4}}} + {1 \over {{x^5}}} - {2 \over {{x^6}}}} \over {{{\left( {1 + {2 \over {{x^3}}}} \right)}^2}}} = 1\)

d. \(\mathop {\lim }\limits_{x \to  + \infty } {{{x^2} + x - 40} \over {2{x^5} + 7{x^4} + 21}} = \mathop {\lim }\limits_{x \to  + \infty } {{{1 \over {{x^3}}} + {1 \over {{x^4}}} - {{40} \over {{x^5}}}} \over {2 + {7 \over x} + {{21} \over {{x^5}}}}} =  + \infty \)

e. Với mọi x < 0, ta có \({1 \over x}\sqrt {2{x^4} + 4{x^2} + 3}  =  - \sqrt {2{x^2} + 4 + {3 \over {{x^2}}}} \)

Do đó :

\(\eqalign{  & \mathop {\lim }\limits_{x \to  - \infty } {{\sqrt {2{x^4} + 4{x^2} + 3} } \over {2x + 1}} = \mathop {\lim }\limits_{x \to  - \infty } {{{1 \over x}\sqrt {2{x^4} + 4{x^2} + 3} } \over {2 + {1 \over x}}}  \cr  &  = \mathop {\lim }\limits_{x \to  - \infty } {{ - \sqrt {2{x^2} + 4 + {3 \over {{x^2}}}} } \over {2 + {1 \over x}}} =  - \infty  \cr} \)

f. \(\mathop {\lim }\limits_{x \to  + \infty } \left( {2x + 1} \right)\sqrt {{{x + 1} \over {2{x^3} + x}}}  = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {{{{{\left( {2x + 1} \right)}^2}\left( {x + 1} \right)} \over {2{x^3} + x}}}  = \sqrt 2 \)

g. \(\mathop {\lim }\limits_{x \to  + \infty } \sqrt {9{x^2} + 11x - 100}  = \mathop {\lim }\limits_{x \to  + \infty } x\sqrt {9 + {{11} \over x} - {{100} \over {{x^2}}}}  =  + \infty \)

h. \(\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {5{x^2} + 1}  - x\sqrt 5 } \right) = \mathop {\lim }\limits_{x \to  + \infty } {1 \over {\sqrt {5{x^2} + 1}  + x\sqrt 5 }} = 0\)

i.

\(\eqalign{  & \mathop {\lim }\limits_{x \to  + \infty } {1 \over {\sqrt {{x^2} + x + 1}  - x}} = \mathop {\lim }\limits_{x \to  + \infty } {{\sqrt {{x^2} + x + 1}  + x} \over {x + 1}}  \cr  &  = \mathop {\lim }\limits_{x \to  + \infty } {{\sqrt {1 + {1 \over x} + {1 \over {{x^2}}}}  + 1} \over {1 + {1 \over x}}} = 2 \cr} \)