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Bài 5. Tính các giới hạn sau
a) \(\mathop {\lim }\limits_{x \to 2} {{x + 3} \over {{x^2} + x + 4}}\)
b) \(\mathop {\lim }\limits_{x \to – 3} {{{x^2} + 5x + 6} \over {{x^2} + 3x}}\)
c) \(\mathop {\lim }\limits_{x \to {4^ – }} {{2x – 5} \over {x – 4}}\)
d) \(\mathop {\lim }\limits_{x \to + \infty } ( – {x^3} + {x^2} – 2x + 1)\)
e) \(\mathop {\lim }\limits_{x \to – \infty } {{x + 3} \over {3x – 1}}\)
f) \(\mathop {\lim }\limits_{x \to – \infty } {{\sqrt {{x^2} – 2x + 4} – x} \over {3x – 1}}\)
a) \(\mathop {\lim }\limits_{x \to 2} {{x + 3} \over {{x^2} + x + 4}} = {{2 + 3} \over {{2^2} + 2 + 4}} = {1 \over 2}\)
b)
\(\eqalign{
& \mathop {\lim }\limits_{x \to – 3} {{{x^2} + 5x + 6} \over {{x^2} + 3x}} = \mathop {\lim }\limits_{x \to – 3} {{(x + 2)(x + 3)} \over {x(x + 3)}} = \mathop {\lim }\limits_{x \to – 3} {{x + 2} \over x} \cr
& = {{ – 3 + 2} \over { – 3}} = {1 \over 3} \cr} \)
c) \(\mathop {\lim }\limits_{x \to {4^ – }} {{2x – 5} \over {x – 4}}\)
Ta có:
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\(\mathop {\lim }\limits_{x \to {4^ – }} (2x – 5) = 3 > 0\)(1)
\(\left\{ \matrix{
x – 4 < 0,\forall x < 4 \hfill \cr
\mathop {\lim }\limits_{x \to – 4} (x – 4) = 0 \hfill \cr} \right.\)
(2)
Từ (1) và (2) suy ra: \(\mathop {\lim }\limits_{x \to {4^ – }} {{2x – 5} \over {x – 4}} = – \infty \)
d) \(\mathop {\lim }\limits_{x \to + \infty } ( – {x^3} + {x^2} – 2x + 1) = \mathop {\lim }\limits_{x \to + \infty } {x^3}( – 1 + {1 \over x} – {2 \over {{x^2}}} + {1 \over {{x^3}}}) = – \infty \)
e)
\(\eqalign{
& \mathop {\lim }\limits_{x \to – \infty } {{x + 3} \over {3x – 1}} = \mathop {\lim }\limits_{x \to – \infty } {{x(1 + {3 \over x})} \over {x(3 – {1 \over x})}} \cr
& = \mathop {\lim }\limits_{x \to – \infty } {{1 + {3 \over x}} \over {3 – {1 \over x}}} = {1 \over 3} \cr} \)
f)
\(\eqalign{
& \mathop {\lim }\limits_{x \to – \infty } {{\sqrt {{x^2} – 2x + 4} – x} \over {3x – 1}} = \mathop {\lim }\limits_{x \to – \infty } {{|x|\sqrt {1 – {2 \over x} + {4 \over {{x^2}}}} – x} \over {3x – 1}} \cr
& \mathop {\lim }\limits_{x \to – \infty } {{ – x\sqrt {1 – {2 \over x} + {4 \over {{x^2}}}} – x} \over {x(3 – {1 \over x})}} = \mathop {\lim }\limits_{x \to – \infty } {{ – \sqrt {1 – {2 \over x} + {4 \over {{x^2}}}} – 1} \over {3 – {1 \over x}}} = {{ – 2} \over 3} \cr} \).