Câu 34. Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to – \infty } \left( {3{x^3} – 5{x^2} + 7} \right)\)
b. \(\mathop {\lim }\limits_{x \to + \infty } \sqrt {2{x^4} – 3x + 12} \)
a. Ta có:
\(\mathop {\lim }\limits_{x \to – \infty } \left( {3{x^3} – 5{x^2} + 7} \right) = \mathop {\lim }\limits_{x \to – \infty } {x^3}\left( {3 – {5 \over x} + {7 \over {{x^3}}}} \right) = – \infty \)
Vì \(\mathop {\lim }\limits_{x \to – \infty } {x^3} = – \infty \,\text{ và }\,\mathop {\lim }\limits_{x \to – \infty } \left( {3 – {5 \over x} + {7 \over {{x^3}}}} \right) = 3 > 0\)
b.
\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } \sqrt {2{x^4} – 3x + 12} = \mathop {\lim }\limits_{x \to + \infty } {x^2}\sqrt {2 – {3 \over {{x^3}}} + {{12} \over {{x^4}}}} = + \infty \cr
& \text{vì }\,\mathop {\lim }\limits_{x \to + \infty } {x^2} = + \infty \,\text{ và }\,\mathop {\lim }\limits_{x \to + \infty } \sqrt {2 – {3 \over {{x^3}}} + {{12} \over {{x^4}}}} = \sqrt 2 > 0 \cr} \)
Câu 35. Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to {2^ + }} {{2x + 1} \over {x – 2}}\)
b. \(\mathop {\lim }\limits_{x \to {2^ – }} {{2x + 1} \over {x – 2}}\)
c. \(\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} – {1 \over {{x^2}}}} \right)\)
d. \(\mathop {\lim }\limits_{x \to {2^ – }} \left( {{1 \over {x – 2}} – {1 \over {{x^2} – 4}}} \right)\)
Advertisements (Quảng cáo)
a.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ + }} {{2x + 1} \over {x – 2}} = + \infty \cr
& \text{vì }\,\mathop {\lim }\limits_{x \to {2^ + }} \left( {2x + 1} \right) = 5,\mathop {\lim }\limits_{x \to {2^ + }} \left( {x – 2} \right) = 0\,\text{ và }\,x – 2 > 0,\forall x > 2 \cr} \)
b.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ – }} {{2x + 1} \over {x – 2}} = – \infty \cr
& \text{vì }\,\mathop {\lim }\limits_{x \to {2^ – }} \left( {2x + 1} \right) = 5,\mathop {\lim }\limits_{x \to {2^ – }} \left( {x – 2} \right) = 0\,\text{ và }\,x – 2 < 0,\forall x < 2 \cr} \)
c.
\(\eqalign{
& \mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} – {1 \over {{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} {{x – 1} \over {{x^2}}} = – \infty \cr
& \text{vì }\,\mathop {\lim }\limits_{x \to 0} \left( {x – 1} \right) = – 1 < 0\,\text{ và }\,\mathop {\lim }\limits_{x \to 0} {x^2} = 0,{x^2} > 0\;\forall x \ne 0. \cr} \)
d.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ – }} \left( {{1 \over {x – 2}} – {1 \over {{x^2} – 4}}} \right) = \mathop {\lim }\limits_{x \to {2^ – }} {{x + 2 – 1} \over {{x^2} – 4}} = \mathop {\lim }\limits_{x \to {2^ – }} {{x + 1} \over {{x^2} – 4}} = – \infty \cr
& \text{vì }\,\mathop {\lim }\limits_{x \to {2^ – }} \left( {x + 1} \right) = 3,\mathop {\lim }\limits_{x \to {2^ – }} \left( {{x^2} – 4} \right) = 0\,\text{ và }\,{x^2} – 4 < 0\,\text{ với }\, – 2 < x < 2 \cr} \)
Câu 36. Tìm các giới hạn sau :
Advertisements (Quảng cáo)
a. \(\mathop {\lim }\limits_{x \to + \infty } {{{x^3} – 5} \over {{x^2} + 1}}\)
b. \(\mathop {\lim }\limits_{x \to – \infty } {{\sqrt {{x^4} – x} } \over {1 – 2x}}\)
a.
\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {{{x^3} – 5} \over {{x^2} + 1}} = \mathop {\lim }\limits_{x \to + \infty } {x}{{{x^2}\left( {1 – {5 \over {{x^3}}}} \right)} \over {{x^2}\left( {1 + {1 \over {{x^2}}}} \right)}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } x.{{1 – {5 \over {{x^3}}}} \over {1 + {1 \over {{x^2}}}}} = + \infty \cr
& \text{vì}\,\mathop {\lim }\limits_{x \to + \infty } x = + \infty \,\text{và}\,\mathop {\lim }\limits_{x \to + \infty } {{1 – {5 \over {{x^3}}}} \over {1 + {1 \over {{x^2}}}}} = 1 > 0 \cr} \)
b.
Với mọi \(x < 0\), ta có \({{\sqrt {{x^4} – x} } \over {1 – 2x}} = {{{x^2}\sqrt {1 – {1 \over {{x^3}}}} } \over {1 – 2x}} = {{\sqrt {1 – {1 \over {{x^3}}}} } \over {{1 \over {{x^2}}} – {2 \over x}}}\)
Vì \(\mathop {\lim }\limits_{x \to – \infty } \sqrt {1 – {1 \over {{x^3}}}} = 1,\mathop {\lim }\limits_{x \to – \infty } \left( {{1 \over {{x^2}}} – {2 \over x}} \right) = 0\,\text{ và }\,{1 \over {{x^2}}} – {2 \over x} > 0\) với mọi \(x < 0\)
Nên \(\mathop {\lim }\limits_{x \to – \infty } {{\sqrt {{x^4} – x} } \over {1 – 2x}} = + \infty \)
Câu 37. a. \(\mathop {\lim }\limits_{x \to 1} \left[ {{2 \over {{{\left( {x – 1} \right)}^2}}}.{{2x + 1} \over {2x – 3}}} \right]\)
b. \(\mathop {\lim }\limits_{x \to 1} {5 \over {\left( {x – 1} \right)\left( {{x^2} – 3x + 2} \right)}}\)
a. Ta có: \(\mathop {\lim }\limits_{x \to 1} {2 \over {{{\left( {x – 1} \right)}^2}}} = + \infty \,\text{ và }\,\mathop {\lim }\limits_{x \to 1} {{2x + 1} \over {2x – 3}} = {3 \over { – 1}} = – 3 < 0\)
Do đó \(\mathop {\lim }\limits_{x \to 1} \left[ {{2 \over {{{\left( {x – 1} \right)}^2}}}.{{2x + 1} \over {2x – 3}}} \right] = – \infty \)
b.
\(\eqalign{
& {5 \over {\left( {x – 1} \right)\left( {{x^2} – 3x + 2} \right)}} = {1 \over {{{\left( {x – 1} \right)}^2}}}.{5 \over {x – 2}} \cr
& \text{vì}\,\mathop {\lim }\limits_{x \to 1} {1 \over {{{\left( {x – 1} \right)}^2}}} = + \infty \,\text{ và }\,\mathop {\lim }\limits_{x \to 1} {5 \over {x – 2}} = – 5 < 0 \cr
& \text{ nên }\,\mathop {\lim }\limits_{x \to 1} {5 \over {\left( {x – 1} \right)\left( {{x^2} – 3x + 2} \right)}} = – \infty \cr} \)
