Giải bài 13 Ôn tập cuối năm

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

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

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

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

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

\(= {{2 - 1} \over {(2 + 2)(2 + \sqrt {3.2 - 2} )}} = {1 \over {16}} \)

c) Ta có:

• \(\mathop {\lim }\limits_{x \to {2^ + }} ({x^2} - 3x + 1) = 4 - 6 + 1 =  - 1\) 
• \(\left\{ \matrix{x - 2 > 0 \hfill \cr \mathop {\lim }\limits_{x \to {2^ + }} (x - 2) = 0 \hfill \cr} \right.\)

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

d) Ta có:

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

\( \left\{ \matrix{1 - x > 0,\forall x < 1 \hfill \cr \mathop {\lim }\limits_{x \to {1^ - }} (1 - x) = 0 \hfill \cr} \right.\)

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

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

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

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

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

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

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

\(= {{1 - \sqrt 4 } \over { - 3}} = {1 \over 3}  \)

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

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

\(= + \infty\)