Bài 13 trang 180 SGK Đại số và Giải tích 11

Tính các giới hạn sau

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

b) \(\mathop {\lim }\limits_{x \to 2} {{x - \sqrt {3x - 2} } \over {{x^2} - 4}}\)

c) \(\mathop {\lim }\limits_{x \to {2^ + }} {{{x^2} - 3x + 1} \over {x - 2}}\)                                                 

d) \(\mathop {\lim }\limits_{x \to {1^ - }} (x + {x^2} + ... + {x^n} - {n \over {1 - x}});n \in {N^*}\)

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

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

g) \(\mathop {\lim }\limits_{x \to  - \infty } ( - 2{x^3} + {x^2} - 3x + 1)\)

a) \(\mathop {\lim }\limits_{x \to  - 2} {{6 - 3x} \over {\sqrt {2{x^2} + 1} }} = {{6 - 3( - 2)} \over {\sqrt {2{{( - 2)}^2} + 1} }} = {{12} \over 3} = 4\)

b)

\(\eqalign{
& \mathop {\lim }\limits_{x \to 2} {{x - \sqrt {3x - 2} } \over {{x^2} - 4}} \cr
& = \mathop {\lim }\limits_{x \to 2} {{(x - \sqrt {x - 2} )(x + \sqrt {3x - 2} )} \over {({x^2} - 4)(x + \sqrt {3x - 2} )}} \cr
& = \mathop {\lim }\limits_{x \to 2} {{{x^2} - 3x + 2} \over {({x^2} - 4)(x + \sqrt {3x - 2} )}} \cr
& = \mathop {\lim }\limits_{x \to 2} {{(x - 2)(x - 1)} \over {(x - 2)(x + 2)(x + \sqrt {3x - 2)} }} \cr
& = \mathop {\lim }\limits_{x \to 2} {{x - 1} \over {(x + 2)(x + \sqrt {3x - 2} )}} \cr
& = {{2 - 1} \over {(2 + 2)(2 + \sqrt {3.2 - 2} )}} = {1 \over {16}} \cr} \)

c) Ta có:

+) \(\mathop {\lim }\limits_{x \to {2^ + }} ({x^2} - 3x + 1) = 4 - 6 + 1 =  - 1\) 

+) \(\left\{ \matrix{
x - 2 > 0 \hfill \cr
\mathop {\lim }\limits_{x \to {2^ + }} (x - 2) = 0 \hfill \cr} \right.\)

Do đó: \(\mathop {\lim }\limits_{x \to {2^ + }} {{{x^2} - 3x + 1} \over {x - 2}} =  - \infty \)

d) Ta có:

\(\eqalign{
& \mathop {\lim }\limits_{x \to 1^-} (x + {x^2} + ... + {x^n} - {n \over {1 - x}}) = - \infty \cr
& \left\{ \matrix{
1 - x > 0,\forall x < 1 \hfill \cr
\mathop {\lim }\limits_{x \to {1^ - }} (1 - x) = 0 \hfill \cr} \right. \cr} \)

+ Suy ra: \(\mathop {\lim }\limits_{x \to {1^ - }} {n \over {1 - x}} =  + \infty \)

+ Do đó: \(\mathop {\lim }\limits_{x \to {1^ - }} (x + {x^2} + ... + {x^n} - {n \over {1 - x}}) =  - \infty \)

e)\(\mathop {\lim }\limits_{x \to  + \infty } {{2x - 1} \over {x + 3}} = \mathop {\lim }\limits_{x \to  + \infty } {{x(2 - {1 \over x})} \over {x(1 + {3 \over x})}} = \mathop {\lim }\limits_{x \to  + \infty } {{2 - {1 \over x}} \over {1 + {3 \over x}}} = 2\)

f)

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

g) 

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