Trung tâm Gia sư Nhân Tài xin giới thiệu chuyên đề bài giảng về phương pháp giải bài tập Giới hạn Vô định – Dạng \( \infty -\infty \). Kính mời các thầy cô và các em học sinh gân xa cùng đọc!
Phụ huynh – Học sinh có nhu cầu cần tìm gia sư dạy kèm toán lớp 11, vui lòng liên hệ: 094.625.1920 – Thầy Nhân (Zalo).
Giới hạn của hàm số – Các dạng giới hạn vô định: (Dạng \( \infty -\infty \))
Dạng tổng quát của dạng này là:
\( \underset{\begin{smallmatrix} x\to \infty \\ \left( x\to {{x}_{O}} \right)\end{smallmatrix}}{\mathop{\lim }}\,\left[ f(x)-g(x) \right] \) trong đó \( \underset{\begin{smallmatrix}x\to \infty \\\left( x\to {{x}_{O}} \right)\end{smallmatrix}}{\mathop{\lim }}\,f(x)=\underset{\begin{smallmatrix} x\to \infty \\ \left( x\to {{x}_{O}} \right)\end{smallmatrix}}{\mathop{\lim }}\,f(x)=\infty \)
Phương pháp: biến đổi chúng về dạng vô định \( \frac{0}{0} \), \( \frac{\infty }{\infty } \)bằng cách biến đổi, nhân liên hợp, thêm bớt, ….
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Hằng đẳng thức |
Liên hợp với a – b |
Liên hợp với a + b |
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\( {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) \) \( {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}}\right)\) \( {{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \) |
a + b |
a – b |
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\( {{a}^{2}}+ab+{{b}^{2}} \) |
\( {{a}^{2}}-ab+{{b}^{2}} \) | |
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A liên hợp với B thì B liên hợp với A |
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Ví dụ 1. Tính \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}-x+1}-\sqrt{{{x}^{2}}+4x+1} \right)\ \)
Phân tích: \( \sqrt{{{\left( +\infty \right)}^{2}}}-\sqrt{{{\left( +\infty \right)}^{2}}}=\infty -\infty \) ( Dạng \( \infty -\infty \))
Phương pháp giải: Dựa vào hằng đẳng thức \( {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) \), ta có liên hợp:
\( \sqrt{A}-\sqrt{B}=\frac{\left( \sqrt{A}-\sqrt{B} \right)\left( \sqrt{A}+\sqrt{B} \right)}{\sqrt{A}+\sqrt{B}}=\frac{A-B}{\sqrt{A}+\sqrt{B}} \)
Giải chi tiết:
\( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}-x+1}-\sqrt{{{x}^{2}}+4x+1} \right) \\ =\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( \sqrt{{{x}^{2}}-x+1}-\sqrt{{{x}^{2}}+4x+1} \right)\left( \sqrt{{{x}^{2}}-x+1}+\sqrt{{{x}^{2}}+4x+1} \right)}{\sqrt{{{x}^{2}}-x+1}+\sqrt{{{x}^{2}}+4x+1}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( {{x}^{2}}-x+1 \right)-\left( {{x}^{2}}+4x+1 \right)}{\sqrt{{{x}^{2}}-x+1}+\sqrt{{{x}^{2}}+4x+1}}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{x}^{2}}-x+1-{{x}^{2}}-4x-1}{\sqrt{{{x}^{2}}-x+1}+\sqrt{{{x}^{2}}+4x+1}} \\ =\underset{x\to +\infty }{\mathop{\lim }}\,\frac{-5x}{\sqrt{{{x}^{2}}-x+1}+\sqrt{{{x}^{2}}+4x+1}}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{-5x}{x\left( \sqrt{1-\frac{1}{x}+\frac{1}{{{x}^{2}}}}+\sqrt{1+\frac{4}{x}+\frac{1}{{{x}^{2}}}} \right)} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{-5}{\sqrt{1-\frac{1}{x}+\frac{1}{{{x}^{2}}}}+\sqrt{1+\frac{4}{x}+\frac{1}{{{x}^{2}}}}}=\frac{-5}{\sqrt{1-0+0}+\sqrt{1+0+0}}=-\frac{5}{2} \\ \)
Vậy \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}-x+1}-\sqrt{{{x}^{2}}+4x+1} \right)=-\frac{5}{2} \)
Ví dụ 2. Tính \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( 2x-1-\sqrt{4{{x}^{2}}-4x-3} \right)\)
Phân tích: \( 2\left( +\infty \right)-\sqrt{4{{\left( +\infty \right)}^{2}}}=\infty -\infty \) (Dạng \( \infty -\infty \))
Phương pháp giải: Dựa vào hằng đẳng thức \( {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) \), ta có:
Liên hợp \( A-\sqrt{B}=\frac{\left( A-\sqrt{B} \right)\left( A+\sqrt{B} \right)}{A+\sqrt{B}}=\frac{{{A}^{2}}-B}{A+\sqrt{B}} \)
Giải chi tiết:
\( \underset{x\to +\infty }{\mathop{\lim }}\,\left( 2x-1-\sqrt{4{{x}^{2}}-4x-3} \right) \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( 2x-1-\sqrt{4{{x}^{2}}-4x-3} \right)\left( 2x-1+\sqrt{4{{x}^{2}}-4x-3} \right)}{2x-1+\sqrt{4{{x}^{2}}-4x-3}} \\ =\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{\left( 2x-1 \right)}^{2}}-{{\left( \sqrt{4{{x}^{2}}-4x-3} \right)}^{2}}}{2x-1+\sqrt{4{{x}^{2}}-4x-3}}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{\left( 2x-1 \right)}^{2}}-\left( 4{{x}^{2}}-4x-3 \right)}{2x-1+\sqrt{4{{x}^{2}}-4x-3}} \\ =\underset{x\to +\infty }{\mathop{\lim }}\,\frac{4{{x}^{2}}-4x+1-4{{x}^{2}}+4x+3}{2x-1+\sqrt{4{{x}^{2}}-4x-3}}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{4}{2x-1+\sqrt{4{{x}^{2}}-4x-3}} \\ =\underset{x\to +\infty }{\mathop{\lim }}\,\frac{x.\frac{4}{x}}{x\left( 2-\frac{1}{x}+\sqrt{4-\frac{4}{x}-\frac{3}{{{x}^{2}}}} \right)}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\frac{4}{x}}{2-\frac{1}{x}+\sqrt{4-\frac{4}{x}-\frac{3}{{{x}^{2}}}}} \\=\frac{0}{2-0+\sqrt{4-0-0}}=0 \\ \)
Vậy \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( 2x-1-\sqrt{4{{x}^{2}}-4x-3} \right)=0\)
Ví dụ 3. Tính \( \underset{x\to +\infty }{\mathop{\lim }}\,x\left( \sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}-2} \right)\)
Phân tích: nếu nhân phân phối \( x\sqrt{{{x}^{2}}+1}-x\sqrt{{{x}^{2}}-2}=\left( +\infty \right)\sqrt{{{\left( +\infty \right)}^{2}}}-\left( +\infty \right)\sqrt{{{\left( +\infty \right)}^{2}}}=\infty -\infty \) (Dạng \( \infty -\infty \))
Phương pháp giải: Dựa vào hằng đẳng thức \( {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) \), ta có liên hợp:
\( \sqrt{A}-\sqrt{B}=\frac{\left( \sqrt{A}-\sqrt{B} \right)\left( \sqrt{A}+\sqrt{B} \right)}{\sqrt{A}+\sqrt{B}}=\frac{A-B}{\sqrt{A}+\sqrt{B}} \)
Giải chi tiết:
\( \underset{x\to +\infty }{\mathop{\lim }}\,x\left( \sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}-2} \right) \\=\underset{x\to +\infty }{\mathop{\lim }}\,x\frac{\left( \sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}-2} \right)\left( \sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}-2} \right)}{\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}-2}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,x\frac{\left( {{x}^{2}}+1 \right)-\left( {{x}^{2}}-2 \right)}{\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}-2}}=\underset{x\to +\infty }{\mathop{\lim }}\,x\frac{{{x}^{2}}+1-{{x}^{2}}+2}{\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}-2}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,x\frac{3}{\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}-2}}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{3x}{x\left( \sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{2}{x}} \right)} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{3}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{2}{x}}}=\frac{3}{\sqrt{1+0}+\sqrt{1-0}}=\frac{3}{2} \\ \)
Vậy \( \underset{x\to +\infty }{\mathop{\lim }}\,x\left( \sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}-2} \right)=\frac{3}{2} \)
Ví dụ 4. Tính \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}+x}-x \right) \)
Hướng dẫn: tương tự nhân và chia biểu thức liên hợp tương ứng là: \( \sqrt{{{x}^{2}}+x}+x \), ta được:
\( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}+x}-x \right)=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( \sqrt{{{x}^{2}}+x}-x \right)\left( \sqrt{{{x}^{2}}+x}+x \right)}{\sqrt{{{x}^{2}}+x}+x} \\ =\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( {{x}^{2}}+x \right)-{{x}^{2}}}{\sqrt{{{x}^{2}}+x}+x}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{x}{\sqrt{{{x}^{2}}+x}+x} \)
Vì \( x\to +\infty \)nên chia cả tử và mẫu cho x ta được:
\( \underset{x\to +\infty }{\mathop{\lim }}\,\frac{x}{\sqrt{{{x}^{2}}+x}+x}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{x}{x\left( \sqrt{1+\frac{1}{x}}+1 \right)} \\ =\underset{x\to +\infty }{\mathop{\lim }}\,\frac{1}{\sqrt{1+\frac{1}{x}}+1}=\frac{1}{\sqrt{1+0}+1}=\frac{1}{2} \)
Vậy: \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}+x}-x \right)=\frac{1}{2} \)
Ví dụ 5. Tính \( \underset{x\to +\infty }{\mathop{\lim }}\,\left[ \sqrt{x+\sqrt{x}}-\sqrt{x} \right] \)
Hướng dẫn giải: Nhân và chia cho biểu thức liên hợp \( \sqrt{x+\sqrt{x}}+\sqrt{x} \)
Giải chi tiết:
\( \underset{x\to +\infty }{\mathop{\lim }}\,\left[ \sqrt{x+\sqrt{x}}-\sqrt{x} \right]=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( \sqrt{x+\sqrt{x}}-\sqrt{x} \right)\left( \sqrt{x+\sqrt{x}}+\sqrt{x} \right)}{\sqrt{x+\sqrt{x}}+\sqrt{x}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( x+\sqrt{x} \right)-x}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\sqrt{x}}{\sqrt{x}\left( \sqrt{1+\frac{1}{\sqrt{x}}}+1 \right)} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{1}{\sqrt{1+\frac{1}{\sqrt{x}}}+1}=\frac{1}{\sqrt{1+0}+1}=\frac{1}{2} \)
Vậy \( \underset{x\to +\infty }{\mathop{\lim }}\,\left[ \sqrt{x+\sqrt{x}}-\sqrt{x} \right]=\frac{1}{2} \)
Ví dụ 6. Tính \( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \sqrt{{{x}^{2}}-x+3}+x \right] \)
Trong ví dụ này cần lưu ý khi \( x\to \infty \)cần xét hai trường hợp \( x\to +\infty \)và \( x\to -\infty \).
+ Khi \( x\to +\infty \) thì:
\( \sqrt{{{x}^{2}}-x+3}x\to +\infty \Rightarrow \sqrt{{{x}^{2}}-x+3}+x\to +\infty \)
Do đó, \( \underset{x\to +\infty }{\mathop{\lim }}\,\left[ \sqrt{{{x}^{2}}-x+3}+x \right]=+\infty \)
+ Khi \( x\to -\infty \)thì:
\( \sqrt{{{x}^{2}}-x+3}x\to +\infty \Rightarrow \sqrt{{{x}^{2}}-x+3}+x=\infty -\infty \\\)
\( \underset{x\to -\infty }{\mathop{\lim }}\,\left[ \sqrt{{{x}^{2}}-x+3}+x \right]=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{\left( \sqrt{{{x}^{2}}-x+3}+x \right)\left( \sqrt{{{x}^{2}}-x+3}-x \right)}{\sqrt{{{x}^{2}}-x+3}-x} \\=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{\left( {{x}^{2}}-x+3 \right)-{{x}^{2}}}{\sqrt{{{x}^{2}}-x+3}-x}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{-x+3}{\sqrt{{{x}^{2}}-x+3}-x}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{x\left( -1+\frac{3}{x} \right)}{\left| x \right|\sqrt{1-\frac{1}{x}+\frac{3}{{{x}^{2}}}}-x} \\=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{x\left( -1+\frac{3}{x} \right)}{-x\sqrt{1-\frac{1}{x}+\frac{3}{{{x}^{2}}}}-x}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{x\left( -1+\frac{3}{x} \right)}{x\left( -\sqrt{1-\frac{1}{x}+\frac{3}{{{x}^{2}}}}-1 \right)} \\=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{-1+\frac{3}{x}}{-\sqrt{1-\frac{1}{x}+\frac{3}{{{x}^{2}}}}-1}=\frac{-1+0}{-\sqrt{1-0+0}-1}=\frac{1}{2} \)
Chú ý: \( \sqrt{{{x}^{2}}}=\left| x \right| \) mà \( x\to -\infty \Rightarrow \left| x \right|=-x \)
Vậy \( \underset{x\to +\infty }{\mathop{\lim }}\,\left[ \sqrt{{{x}^{2}}-x+3}+x \right]=+\ \)infty và \( \underset{x\to -\infty }{\mathop{\lim }}\,\left[ \sqrt{{{x}^{2}}-x+3}+x \right]=\frac{1}{2} \)
Ví dụ 7. Tính \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt[3]{{{x}^{3}}+5{{x}^{2}}}-\sqrt[3]{{{x}^{3}}+8x} \right)\)
Phân tích: \( \sqrt[3]{{{\left( +\infty \right)}^{3}}}-\sqrt[3]{{{\left( +\infty \right)}^{3}}}=\infty -\infty \)
Phương pháp giải: Dựa vào hằng đẳng thức \( {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) \), ta có liên hợp:
\( \sqrt[3]{A}-\sqrt[3]{B}=\frac{\left( \sqrt[3]{A}-\sqrt[3]{B} \right)\left[ {{\left( \sqrt[3]{A} \right)}^{2}}+\sqrt[3]{A}.\sqrt[3]{B}+{{\left( \sqrt[3]{B} \right)}^{2}} \right]}{{{\left( \sqrt[3]{A} \right)}^{2}}+\sqrt[3]{A}.\sqrt[3]{B}+{{\left( \sqrt[3]{B} \right)}^{2}}}=\frac{A-B}{{{\left( \sqrt[3]{A} \right)}^{2}}+\sqrt[3]{A}.\sqrt[3]{B}+{{\left( \sqrt[3]{B} \right)}^{2}}} \)
Giải chi tiết:
\( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt[3]{{{x}^{3}}+5{{x}^{2}}}-\sqrt[3]{{{x}^{3}}+8x} \right) \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( \sqrt[3]{{{x}^{3}}+5{{x}^{2}}}-\sqrt[3]{{{x}^{3}}+8x} \right)\left[ {{\left( \sqrt[3]{{{x}^{3}}+5{{x}^{2}}} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+5{{x}^{2}}}.\sqrt[3]{{{x}^{3}}+8x}+{{\left( \sqrt[3]{{{x}^{3}}+8x} \right)}^{2}} \right]}{{{\left( \sqrt[3]{{{x}^{3}}+5{{x}^{2}}} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+5{{x}^{2}}}.\sqrt[3]{{{x}^{3}}+8x}+{{\left( \sqrt[3]{{{x}^{3}}+8x} \right)}^{2}}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( {{x}^{3}}+5{{x}^{2}} \right)-\left( {{x}^{3}}+8x \right)}{{{\left( \sqrt[3]{{{x}^{3}}+5{{x}^{2}}} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+5{{x}^{2}}}.\sqrt[3]{{{x}^{3}}+8x}+{{\left( \sqrt[3]{{{x}^{3}}+8x} \right)}^{2}}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{x}^{3}}+5{{x}^{2}}-{{x}^{3}}-8x}{{{\left( \sqrt[3]{{{x}^{3}}+5{{x}^{2}}} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+5{{x}^{2}}}.\sqrt[3]{{{x}^{3}}+8x}+{{\left( \sqrt[3]{{{x}^{3}}+8x} \right)}^{2}}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{5{{x}^{2}}-8x}{{{\left( \sqrt[3]{{{x}^{3}}+5{{x}^{2}}} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+5{{x}^{2}}}.\sqrt[3]{{{x}^{3}}+8x}+{{\left( \sqrt[3]{{{x}^{3}}+8x} \right)}^{2}}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{x}^{2}}\left( 5-\frac{8}{x} \right)}{{{x}^{2}}\left[ {{\left( \sqrt[3]{1+\frac{5}{x}} \right)}^{2}}+\sqrt[3]{1+\frac{5}{x}}.\sqrt[3]{1+\frac{8}{x}}+{{\left( \sqrt[3]{1+\frac{8}{x}} \right)}^{2}} \right]} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{5-\frac{8}{x}}{{{\left( \sqrt[3]{1+\frac{5}{x}} \right)}^{2}}+\sqrt[3]{1+\frac{5}{x}}.\sqrt[3]{1+\frac{8}{x}}+{{\left( \sqrt[3]{1+\frac{8}{x}} \right)}^{2}}} \\=\frac{5-0}{{{\left( \sqrt[3]{1+0} \right)}^{2}}+\sqrt[3]{1+0}.\sqrt[3]{1+0}+{{\left( \sqrt[3]{1+0} \right)}^{2}}}=\frac{5}{3} \)
Vậy \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt[3]{{{x}^{3}}+5{{x}^{2}}}-\sqrt[3]{{{x}^{3}}+8x} \right)=\frac{5}{3}\)
Ví dụ 8. Tính \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( x+\sqrt[3]{3{{x}^{2}}-{{x}^{3}}} \right) \)
Phân tích: \( \left( +\infty \right)+\sqrt[3]{-{{\left( +\infty \right)}^{3}}}=\infty -\infty \)
Phương pháp giải: Dựa vào hằng đẳng thức \( {{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \), ta có liên hợp:
\( A+\sqrt[3]{B}=\frac{\left( A+\sqrt[3]{B} \right)\left[ {{\left( A \right)}^{2}}-A.\sqrt[3]{B}+{{\left( \sqrt[3]{B} \right)}^{2}} \right]}{{{\left( A \right)}^{2}}-A.\sqrt[3]{B}+{{\left( \sqrt[3]{B} \right)}^{2}}}=\frac{{{A}^{3}}+B}{{{\left( A \right)}^{2}}-A.\sqrt[3]{B}+{{\left( \sqrt[3]{B} \right)}^{2}}} \)
Giải chi tiết:
\( \underset{x\to +\infty }{\mathop{\lim }}\,\left( x+\sqrt[3]{3{{x}^{2}}-{{x}^{3}}} \right) \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( x+\sqrt[3]{3{{x}^{2}}-{{x}^{3}}} \right)\left[ {{x}^{2}}-x.\sqrt[3]{3{{x}^{2}}-{{x}^{3}}}+{{\left( \sqrt[3]{3{{x}^{2}}-{{x}^{3}}} \right)}^{2}} \right]}{{{x}^{2}}-x.\sqrt[3]{3{{x}^{2}}-{{x}^{3}}}+{{\left( \sqrt[3]{3{{x}^{2}}-{{x}^{3}}} \right)}^{2}}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{x}^{3}}+\left( 3{{x}^{2}}-{{x}^{3}} \right)}{{{x}^{2}}-x.\sqrt[3]{3{{x}^{2}}-{{x}^{3}}}+{{\left( \sqrt[3]{3{{x}^{2}}-{{x}^{3}}} \right)}^{2}}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{x}^{3}}+3{{x}^{2}}-{{x}^{3}}}{{{x}^{2}}-x.\sqrt[3]{3{{x}^{2}}-{{x}^{3}}}+{{\left( \sqrt[3]{3{{x}^{2}}-{{x}^{3}}} \right)}^{2}}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{3{{x}^{2}}}{{{x}^{2}}\left[ 1-\sqrt[3]{\frac{3}{x}-1}+{{\left( \sqrt[3]{\frac{3}{x}-1} \right)}^{2}} \right]} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{3}{1-\sqrt[3]{\frac{3}{x}-1}+{{\left( \sqrt[3]{\frac{3}{x}-1} \right)}^{2}}}=\frac{3}{1-\sqrt[3]{0-1}+{{\left( \sqrt[3]{0-1} \right)}^{2}}}=1 \)
Vậy \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( x+\sqrt[3]{3{{x}^{2}}-{{x}^{3}}} \right)=1 \)
Ví dụ 9. Tính \( \underset{x\to +\infty }{\mathop{\lim }}\,{{x}^{2}}.\left( \sqrt[3]{{{x}^{3}}+1}-x \right) \)
Phân tích: Nếu nhân phân phối vào thì \( {{x}^{2}}.\left( \sqrt[3]{{{x}^{3}}+1}-x \right)={{x}^{2}}\sqrt[3]{{{x}^{3}}+1}-{{x}^{3}}={{\left( +\infty \right)}^{2}}\sqrt[3]{{{\left( +\infty \right)}^{3}}}-{{\left( +\infty \right)}^{3}}=\infty -\infty \)
Phương pháp giải: Dựa vào hằng đẳng thức \( {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) \), ta có liên hợp:
\( \sqrt[3]{A}-B=\frac{\left( \sqrt[3]{A}-B \right)\left[ {{\left( \sqrt[3]{A} \right)}^{2}}+\sqrt[3]{A}.B+{{\left( B \right)}^{2}} \right]}{{{\left( \sqrt[3]{A} \right)}^{2}}+\sqrt[3]{A}.B+{{\left( B \right)}^{2}}}=\frac{A-{{B}^{3}}}{{{\left( \sqrt[3]{A} \right)}^{2}}+\sqrt[3]{A}.B+{{\left( B \right)}^{2}}} \)
Giải chi tiết:
\( \underset{x\to +\infty }{\mathop{\lim }}\,{{x}^{2}}.\left( \sqrt[3]{{{x}^{3}}+1}-x \right) \\=\underset{x\to +\infty }{\mathop{\lim }}\,{{x}^{2}}.\frac{\left( \sqrt[3]{{{x}^{3}}+1}-x \right)\left[ {{\left( \sqrt[3]{{{x}^{3}}+1} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+1}.x+{{x}^{2}} \right]}{{{\left( \sqrt[3]{{{x}^{3}}+1} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+1}.x+{{x}^{2}}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,{{x}^{2}}.\frac{\left( {{x}^{3}}+1 \right)-{{x}^{3}}}{{{\left( \sqrt[3]{{{x}^{3}}+1} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+1}.x+{{x}^{2}}}=\underset{x\to +\infty }{\mathop{\lim }}\,{{x}^{2}}.\frac{1}{{{\left( \sqrt[3]{{{x}^{3}}+1} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+1}.x+{{x}^{2}}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{x}^{2}}}{{{x}^{2}}\left[ {{\left( \sqrt[3]{1+\frac{1}{{{x}^{3}}}} \right)}^{2}}+\sqrt[3]{1+\frac{1}{{{x}^{3}}}}+1 \right]}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{1}{{{\left( \sqrt[3]{1+\frac{1}{{{x}^{3}}}} \right)}^{2}}+\sqrt[3]{1+\frac{1}{{{x}^{3}}}}+1} \\ =\frac{1}{{{\left( \sqrt[3]{1+0} \right)}^{2}}+\sqrt[3]{1+0}+1}=\frac{1}{3}\)
Vậy \( \underset{x\to +\infty }{\mathop{\lim }}\,{{x}^{2}}.\left( \sqrt[3]{{{x}^{3}}+1}-x \right)=\frac{1}{3} \)
Ví dụ 10. Tính \( \underset{x\to +\infty }{\mathop{\lim }}\,\left[ \sqrt[3]{{{x}^{3}}+3{{x}^{2}}}-\sqrt{{{x}^{2}}-2x} \right]\)
Hướng dẫn:
Vì hàm số cần tìm giới hạn chứa các căn thức không cùng bậc nên ta thêm bớt để có thể nhân liên hợp
Giải chi tiết:
\( \underset{x\to +\infty }{\mathop{\lim }}\,\left[ \sqrt[3]{{{x}^{3}}+3{{x}^{2}}}-\sqrt{{{x}^{2}}-2x} \right]=\underset{x\to +\infty }{\mathop{\lim }}\,\left[ \left( \sqrt[3]{{{x}^{3}}+3{{x}^{2}}}-x \right)-\left( \sqrt{{{x}^{2}}-2{{x}^{2}}}-x \right) \right] \\=\underset{x\to +\infty }{\mathop{\lim }}\,\left[ \sqrt[3]{{{x}^{3}}+3{{x}^{2}}}-x \right]-\underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}-2x}-x \right)={{L}_{1}}-{{L}_{2}} \\ +)\text{ }{{L}_{1}}=\underset{x\to +\infty }{\mathop{\lim }}\,\left[ \sqrt[3]{{{x}^{3}}+3{{x}^{2}}}-x \right]=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( \sqrt[3]{{{x}^{3}}+3{{x}^{2}}}-x \right)\left[ {{\left( \sqrt[3]{{{x}^{3}}+3{{x}^{2}}} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+3{{x}^{2}}}.x+{{x}^{2}} \right]}{{{\left( \sqrt[3]{{{x}^{3}}+3{{x}^{2}}} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+3{{x}^{2}}}.x+{{x}^{2}}} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{x}^{3}}+3{{x}^{2}}-{{x}^{3}}}{{{\left( \sqrt[3]{{{x}^{3}}+3{{x}^{2}}} \right)}^{2}}+\sqrt[3]{{{x}^{3}}+3{{x}^{2}}}.x+{{x}^{2}}}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{3{{x}^{2}}}{{{x}^{2}}\left[ {{\left( \sqrt[3]{1+\frac{3}{x}} \right)}^{2}}+\sqrt[3]{1+\frac{3}{x}}+1 \right]} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{3}{{{\left( \sqrt[3]{1+\frac{3}{x}} \right)}^{2}}+\sqrt[3]{1+\frac{3}{x}}+1}=\frac{3}{{{\left( \sqrt[3]{1+0} \right)}^{2}}+\sqrt[3]{1+0}+1}=1 \\+)\text{ }{{L}_{2}}=\underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}-2x}-x \right)=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( \sqrt{{{x}^{2}}-2x}-x \right)\left( \sqrt{{{x}^{2}}-2x}+x \right)}{\sqrt{{{x}^{2}}-2x}+x} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{x}^{2}}-2x-{{x}^{2}}}{\sqrt{{{x}^{2}}-2x}+x}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{-2x}{\sqrt{{{x}^{2}}-2x}+x}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{-2x}{x\left( \sqrt{1-\frac{2}{x}}+1 \right)} \\=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{-2}{\sqrt{1-\frac{2}{x}}+1}=\frac{-2}{\sqrt{1-0}+1}=-1 \)
Vậy \( \underset{x\to +\infty }{\mathop{\lim }}\,\left[ \sqrt[3]{{{x}^{3}}+3{{x}^{2}}}-\sqrt{{{x}^{2}}-2x} \right]={{L}_{1}}-{{L}_{2}}=1+1=2 \)
Bài tập luyện tập:
1. \( \underset{x\to -\infty }{\mathop{\lim }}\,\left( 2x+\sqrt{4{{x}^{2}}+3x-2} \right) \) (Đs: \( -\frac{3}{4} \))
2. \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt{9{{x}^{2}}-3x-1}-3x \right) \) (Đs: \( \frac{1}{2} \))
3. \( \underset{x\to -\infty }{\mathop{\lim }}\,\left( x\sqrt{3}+\sqrt{3{{x}^{2}}+4x-1} \right) \) (Đs: \( \frac{-2}{\sqrt{3}} \))
4. \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}-2x+3}-x+1 \right) \) (Đs: 0)
5. \( \underset{x\to +\infty }{\mathop{\lim }}\,\left( \sqrt[3]{{{x}^{3}}+{{x}^{2}}+x}-\sqrt{{{x}^{2}}+x+1} \right) \) (Đs: \( -\frac{1}{6} \))
6. \( \underset{x\to -\infty }{\mathop{\lim }}\,\left( \sqrt[3]{{{x}^{3}}+2x-1}+\sqrt{{{x}^{2}}-3} \right) \) (Đs: 0)