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Bài 13. 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:
+) \(\mathop {\lim }\limits_{x \to {2^ + }} ({x^2} – 3x + 1) = 4 – 6 + 1 = – 1\)
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+) \(\left\{ \matrix{
x – 2 > 0 \hfill \cr
\mathop {\lim }\limits_{x \to {2^ + }} (x – 2) = 0 \hfill \cr} \right.\)
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)
\(\eqalign{
& \mathop {\lim }\limits_{x \to – \infty } ( – 2{x^3} + {x^2} – 3x + 1) \cr
& = \mathop {\lim }\limits_{x \to – \infty } {x^3}( – 2 + {1 \over x} – {3 \over {{x^2}}} + {1 \over {{x^3}}}) = + \infty \cr}\)