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