Câu 26. Áp dụng định nghĩa giới hạn bên phải và giới hạn bên trái của hàm số, tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to {1^ + }} \sqrt {x – 1} \)
b. \(\mathop {\lim }\limits_{x \to {5^ – }} \left( {\sqrt {5 – x} + 2x} \right)\)
c. \(\mathop {\lim }\limits_{x \to {3^ + }} {1 \over {x – 3}}\)
d. \(\mathop {\lim }\limits_{x \to {3^ – }} {1 \over {x – 3}}\)
a. \(\mathop {\lim }\limits_{x \to {1^ + }} \sqrt {x – 1} = 0\)
b. \(\mathop {\lim }\limits_{x \to 5} \left( {\sqrt {5 – x} + 2x} \right) = 2.5 = 10\)
c. \(\mathop {\lim }\limits_{x \to {3^ + }} {1 \over {x – 3}} = + \infty \,\left( {\text{ vì }\,x > 3} \right)\)
d. \(\mathop {\lim }\limits_{x \to {3^ – }} {1 \over {x – 3}} = – \infty \,\left( {\text{ vì }\,x < 3} \right)\)
Câu 27. Tìm các giới hạn sau (nếu có) :
a. \(\mathop {\lim }\limits_{x \to {2^ + }} {{\left| {x – 2} \right|} \over {x – 2}}\)
b. \(\mathop {\lim }\limits_{x \to {2^ – }} {{\left| {x – 2} \right|} \over {x – 2}}\)
c. \(\mathop {\lim }\limits_{x \to 2} {{\left| {x – 2} \right|} \over {x – 2}}\)
a. Với mọi \(x > 2\), ta có \(\left| {x – 2} \right| = x – 2.\) Do đó :
\(\mathop {\lim }\limits_{x \to {2^ + }} {{\left| {x – 2} \right|} \over {x – 2}} = \mathop {\lim }\limits_{x \to {2^ + }} {{x – 2} \over {x – 2}} = \mathop {\lim }\limits_{x \to {2^ + }} 1 = 1\)
b. Với mọi \(x < 2\), ta có \(|x – 2| = 2 – x\). Do đó :
\(\mathop {\lim }\limits_{x \to {2^ – }} {{\left| {x – 2} \right|} \over {x – 2}} = \mathop {\lim }\limits_{x \to {2^ – }} {{2 – x} \over {x – 2}} = \mathop {\lim }\limits_{x \to {2^ – }} – 1 = – 1\)
c. Vì \(\mathop {\lim }\limits_{x \to {2^ + }} {{\left| {x – 2} \right|} \over {x – 2}} \ne \mathop {\lim }\limits_{x \to {2^ – }} {{\left| {x – 2} \right|} \over {x – 2}}\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 2} {{\left| {x – 2} \right|} \over {x – 2}}\)
Câu 28. Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to {0^ + }} {{x + 2\sqrt x } \over {x – \sqrt x }}\)
b. \(\mathop {\lim }\limits_{x \to {2^ – }} {{4 – {x^2}} \over {\sqrt {2 – x} }}\)
c. \(\mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} {{{x^2} + 3x + 2} \over {\sqrt {{x^5} + {x^4}} }}\)
d. \(\mathop {\lim }\limits_{x \to {3^ – }} {{\sqrt {{x^2} – 7x + 12} } \over {\sqrt {9 – {x^2}} }}\)
a. Với \(x > 0\), ta có : \({{x + 2\sqrt x } \over {x – \sqrt x }} = {{\sqrt x \left( \sqrt x + 2 \right)} \over {\sqrt x \left( {\sqrt x – 1} \right)}} = {{\sqrt x + 2} \over {\sqrt x – 1}}\)
do đó : \(\mathop {\lim }\limits_{x \to {0^ + }} {{x + 2\sqrt x } \over {x – \sqrt x }} = \mathop {\lim }\limits_{x \to {0^ + }} {{\sqrt x + 2} \over {\sqrt x – 1}} = {2 \over { – 1}} = – 2\)
b. Với \(x < 2\), ta có : \({{4 – {x^2}} \over {\sqrt {2 – x} }} = {{\left( {2 – x} \right)\left( {2 + x} \right)} \over {\sqrt {2 – x} }} = \left( {x + 2} \right)\sqrt {2 – x} \)
Do đó \(\mathop {\lim }\limits_{x \to {2^ – }} {{4 – {x^2}} \over {\sqrt {2 – x} }} = \mathop {\lim }\limits_{x \to {2^ – }} \left( {x + 2} \right)\sqrt {2 – x} = 0\)
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c. Với mọi \(x > -1\)
\({{{x^2} + 3x + 2} \over {\sqrt {{x^5} + {x^4}} }} = {{\left( {x + 1} \right)\left( {x + 2} \right)} \over {{x^2}\sqrt {x + 1} }} = {{\sqrt {x + 1} \left( {x + 2} \right)} \over {{x^2}}}\)
Do đó \(\mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} {{{x^2} + 3x + 2} \over {\sqrt {{x^5} + {x^4}} }} = \mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} {{\sqrt {x + 1} \left( {x + 2} \right)} \over {{x^2}}} = 0\)
d. Với \(-3 < x < 3\)
\({{\sqrt {{x^2} – 7x + 12} } \over {\sqrt {9 – {x^2}} }} = {{\sqrt {\left( {3 – x} \right)\left( {4 – x} \right)} } \over {\sqrt {\left( {3 – x} \right)\left( {3 + x} \right)} }} = {{\sqrt {4 – x} } \over {\sqrt {3 + x} }}\)
Do đó \(\mathop {\lim }\limits_{x \to {3^ – }} {{\sqrt {{x^2} – 7x + 12} } \over {\sqrt {9 – {x^2}} }} = {1 \over {\sqrt 6 }} = {{\sqrt 6 } \over 6}\)
Câu 29. Cho hàm số
\(f\left( x \right) = \left\{ {\matrix{{2\left| x \right| – 1\,\text{ với }\,x \le – 2,} \cr {\sqrt {2{x^2} + 1} \,\text{ với }\,x > – 2.} \cr} } \right.\)
Tìm \(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ – }} f\left( x \right),\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} f\left( x \right)\,\text{ và }\,\mathop {\lim }\limits_{x \to – 2} f\left( x \right)\) (nếu có).
Ta có:
\(\eqalign{
& \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ – }} \left( {2\left| x \right| – 1} \right) = 2\left| { – 2} \right| – 1 = 3 \cr
& \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} = \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} \sqrt {2{x^2} + 1} = 3 \Rightarrow \mathop {\lim }\limits_{x \to – 2} f\left( x \right) = 3. \cr} \)
Câu 30. Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} – 8} \right|\)
b. \(\mathop {\lim }\limits_{x \to 2} {{{x^2} + x + 1} \over {{x^2} + 2x}}\)
c. \(\mathop {\lim }\limits_{x \to – 1} \sqrt {{{{x^3}} \over {{x^2} – 3}}} \)
d. \(\mathop {\lim }\limits_{x \to 3} \root 3 \of {{{2x\left( {x + 1} \right)} \over {{x^2} – 6}}} \)
e. \(\mathop {\lim }\limits_{x \to – 2} {{\sqrt {1 – {x^3}} – 3x} \over {2{x^2} + x – 3}}\)
f. \(\mathop {\lim }\limits_{x \to – 2} {{2\left| {x + 1} \right| – 5\sqrt {{x^2} – 3} } \over {2x + 3}}\)
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a. \(\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} – 8} \right| = \left| {{{\left( {\sqrt 3 } \right)}^2} – 8} \right| = 5\)
b. \(\mathop {\lim }\limits_{x \to 2} {{{x^2} + x + 1} \over {{x^2} + 2x}} = {{{2^2} + 2 + 1} \over {{2^2} + 2.2}} = {7 \over 8}\)
c. \(\mathop {\lim }\limits_{x \to – 1} \sqrt {{{{x^3}} \over {{x^2} – 3}}} = \sqrt {{1 \over 2}} = {{\sqrt 2 } \over 2}\)
d. \(\mathop {\lim }\limits_{x \to 3} \root 3 \of {{{2x\left( {x + 1} \right)} \over {{x^2} – 6}}} = \root 3 \of {{{24} \over 3}} = 2\)
e. \(\mathop {\lim }\limits_{x \to – 2} {{\sqrt {1 – {x^3}} – 3x} \over {2{x^2} + x – 3}} = {{3 + 6} \over {8 – 5}} = 3\)
f. \(\mathop {\lim }\limits_{x \to – 2} {{2\left| {x + 1} \right| – 5\sqrt {{x^2} – 3} } \over {2x + 3}} = {{2 – 5} \over { – 4 + 3}} = 3\)
Câu 31. Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to – \sqrt 2 } {{{x^3} + 2\sqrt 2 } \over {{x^2} – 2}}\)
b. \(\mathop {\lim }\limits_{x \to 3} {{{x^4} – 27x} \over {2{x^2} – 3x – 9}}\)
c. \(\mathop {\lim }\limits_{x \to – 2} {{{x^4} – 16} \over {{x^2} + 6x + 8}}\)
d. \(\mathop {\lim }\limits_{x \to {1^ – }} {{\sqrt {1 – x} + x – 1} \over {\sqrt {{x^2} – {x^3}} }}\)
a. Ta có:
\(\eqalign{
& \mathop {\lim }\limits_{x \to – \sqrt 2 } = {{{x^3} + 2\sqrt 2 } \over {{x^2} – 2}} = \mathop {\lim }\limits_{x \to – \sqrt 2 } {{{x^3} + {{\left( {\sqrt 2 } \right)}^3}} \over {{x^2} – {{\left( {\sqrt 2 } \right)}^2}}} \cr
& = \mathop {\lim }\limits_{x \to – \sqrt 2 } {{\left( {x + \sqrt 2 } \right)\left( {{x^2} – x\sqrt 2 + 2} \right)} \over {\left( {x + \sqrt 2 } \right)\left( {x – \sqrt 2 } \right)}} \cr
& = \mathop {\lim }\limits_{x \to – \sqrt 2 } {{{x^2} – x\sqrt 2 + 2} \over {x – \sqrt 2 }} = {{ – 3\sqrt 2 } \over 2} \cr} \)
b.
\(\eqalign{
& \mathop {\lim }\limits_{x \to 3} {{{x^4} – 27x} \over {2{x^2} – 3x – 9}} = \mathop {\lim }\limits_{x \to 3} {{x\left( {x – 3} \right)\left( {{x^2} + 3x + 9} \right)} \over {\left( {x – 3} \right)\left( {2x + 3} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 3} {{x\left( {{x^2} + 3x + 9} \right)} \over {2x + 3}} = 9 \cr} \)
c.
\(\eqalign{
& \mathop {\lim }\limits_{x \to – 2} {{{x^4} – 16} \over {{x^2} + 6x + 8}} = \mathop {\lim }\limits_{x \to – 2} {{\left( {{x^2} – 4} \right)\left( {{x^2} + 4} \right)} \over {\left( {x + 2} \right)\left( {x + 4} \right)}} \cr
& = \mathop {\lim }\limits_{x \to – 2} {{\left( {x – 2} \right)\left( {{x^2} + 4} \right)} \over {x + 4}} = – 16 \cr} \)
d.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {1^ – }} {{\sqrt {1 – x} + x – 1} \over {\sqrt {{x^2} – {x^3}} }} = \mathop {\lim }\limits_{x \to {1^ – }} {{\sqrt {1 – x} – \left( {1 – x} \right)} \over {\left| x \right|\sqrt {1 – x} }} \cr
& = \mathop {\lim }\limits_{x \to {1^ – }} {{1 – \sqrt {1 – x} } \over {\left| x \right|}} = 1 \cr} \)
Câu 32. Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to + \infty } \root 3 \of {{{2{x^5} + {x^3} – 1} \over {\left( {2{x^2} – 1} \right)\left( {{x^3} + x} \right)}}} \)
b. \(\mathop {\lim }\limits_{x \to – \infty } {{2\left| x \right| + 3} \over {\sqrt {{x^2} + x + 5} }}\)
c. \(\mathop {\lim }\limits_{x \to – \infty } {{\sqrt {{x^2} + x} + 2x} \over {2x + 3}}\)
d. \(\mathop {\lim }\limits_{x \to + \infty } \left( {x + 1} \right)\sqrt {{x \over {2{x^4} + {x^2} + 1}}} \)
a. \(\mathop {\lim }\limits_{x \to + \infty } \root 3 \of {{{2{x^5} + {x^3} – 1} \over {\left( {2{x^2} – 1} \right)\left( {{x^3} + x} \right)}}} = \mathop {\lim }\limits_{x \to + \infty } \root 3 \of {{{2 + {1 \over {{x^2}}} – {1 \over {{x^5}}}} \over {\left( {2 – {1 \over {{x^2}}}} \right)\left( {1 + {1 \over {{x^2}}}} \right)}}} = 1\)
b.
\(\eqalign{
& \mathop {\lim }\limits_{x \to – \infty } {{2\left| x \right| + 3} \over {\sqrt {{x^2} + x + 5} }} = \mathop {\lim }\limits_{x \to – \infty } {{2\left| x \right| + 3} \over {\left| x \right|\sqrt {1 + {1 \over x} + {5 \over {{x^2}}}} }} \cr
& = \mathop {\lim }\limits_{x \to – \infty } {{ – 2x + 3} \over { – x\sqrt {1 + {1 \over x} + {5 \over {{x^2}}}} }} =\mathop {\lim }\limits_{x \to – \infty } {{2 – {3 \over x}} \over {\sqrt {1 + {1 \over x} + {5 \over {{x^2}}}} }}= 2 \cr} \)
c. \({x^2} + x \ge 0 \Leftrightarrow x \le – 1\,\text{ hoặc }\,x \ge 0\)
Với mọi \(x ≤ -1\), \(x \ne – {3 \over 2}\)
\({{\sqrt {{x^2} + x} + 2x} \over {2x + 3}} = {{\left| x \right|\sqrt {1 + {1 \over x}} + 2x} \over {2x + 3}} = {{ – x\sqrt {1+ {1 \over x}} + 2x} \over {2x + 3}} = {{ – \sqrt {1 + {1 \over x}} + 2} \over {2 + {3 \over x}}}\)
Do đó \(\mathop {\lim }\limits_{x \to – \infty } {{\sqrt {{x^2} + x} + 2x} \over {2x + 3}} =\mathop {\lim }\limits_{x \to – \infty }{{ – \sqrt {1 + {1 \over x}} + 2} \over {2 + {3 \over x}}}= {1 \over 2}\)
d.
\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } \left( {x + 1} \right)\sqrt {{x \over {2{x^4} + {x^2} + 1}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {{{x{{\left( {x + 1} \right)}^2}} \over {2{x^4} + {x^2} + 1}}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } \sqrt {{{{1 \over x} + {2 \over {{x^2}}} + {1 \over {{x^3}}}} \over {2 + {1 \over {{x^2}}} + {1 \over {{x^4}}}}}} = 0 \cr} \)
Câu 33. Cho hàm số
\(f\left( x \right) = \left\{ {\matrix{{{x^2} – 2x + 3\,\text{ với }\,x \le 2.} \cr {4x – 3\,\text{ với }\,x > 2} \cr} } \right.\)
Tìm \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right),\mathop {\lim }\limits_{x \to {2^ – }} f\left( x \right)\,\text{ và }\,\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) (nếu có).
Ta có:
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {4x – 3} \right) =4.2-3= 5 \cr
& \mathop {\lim }\limits_{x \to {2^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ – }} \left( {{x^2} – 2x + 3} \right) =2^2-2.2+3= 3 \cr} \)
Vì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^ – }} f\left( x \right)\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\)
