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

##### Hướng dẫn giải

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}