Bài tập nâng cao · Bài 23
Bài tập nâng cao Giới hạn của hàm số · Bài 23
Tìm giới hạn
a. $\lim\limits_{x\to +\infty}\frac{\left( x-1 \right)\sqrt{x^{2}+x\sqrt{x}+1}}{\sqrt{x^{2}+1}+2x}$
b. $\lim\limits_{x\to -\infty}\frac{\left( x-1 \right)\left( 2x-1 \right)\left( 3x-1 \right)\left( 4x-1 \right)\left( 5x-1 \right)}{\left( 4x+5 \right)^{5}}$
c. $\lim\limits_{x\to -\infty}\frac{\left( x-1 \right)\sqrt{x^{2}+x\sqrt{x}+1}}{\sqrt{x^{2}+1}+2x}$
d. $\lim\limits_{x\to +\infty}x^{2}\left( \sqrt[3]{x^{3}+1}-x \right)$
a. $\lim\limits_{x\to +\infty}\frac{\left( x-1 \right)\sqrt{x^{2}+x\sqrt{x}+1}}{\sqrt{x^{2}+1}+2x}$
b. $\lim\limits_{x\to -\infty}\frac{\left( x-1 \right)\left( 2x-1 \right)\left( 3x-1 \right)\left( 4x-1 \right)\left( 5x-1 \right)}{\left( 4x+5 \right)^{5}}$
c. $\lim\limits_{x\to -\infty}\frac{\left( x-1 \right)\sqrt{x^{2}+x\sqrt{x}+1}}{\sqrt{x^{2}+1}+2x}$
d. $\lim\limits_{x\to +\infty}x^{2}\left( \sqrt[3]{x^{3}+1}-x \right)$
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Lời giải
a. $\lim\limits_{x\to +\infty}\frac{\left( x-1 \right)\sqrt{x^{2}+x\sqrt{x}+1}}{\sqrt{x^{2}+1}+2x}=\lim\limits_{x\to +\infty}\frac{\left( x-1 \right)\left| x \right|\sqrt{1+\sqrt{\frac{1}{x}}+\frac{1}{x^{2}}}}{\left| x \right|\sqrt{1+\frac{1}{x^{2}}}+2x}=\lim\limits_{x\to +\infty}\frac{\left( x-1 \right)\sqrt{1+\sqrt{\frac{1}{x}}+\frac{1}{x^{2}}}}{\sqrt{1+\frac{1}{x^{2}}}+2}=+\infty$
b. $\lim\limits_{x\to -\infty}\frac{\left( x-1 \right)\left( 2x-1 \right)\left( 3x-1 \right)\left( 4x-1 \right)\left( 5x-1 \right)}{\left( 4x+5 \right)^{5}}=\lim\limits_{x\to -\infty}\frac{\left( 1-\frac{1}{x} \right)\left( 2-\frac{1}{x} \right)\left( 3-\frac{1}{x} \right)\left( 4-\frac{1}{x} \right)\left( 5-\frac{1}{x} \right)}{\left( 4+\frac{5}{x} \right)^{5}}=\frac{15}{128}$
c. $\lim\limits_{x\to -\infty}\frac{\left( x-1 \right)\sqrt{x^{2}+x\sqrt{x}+1}}{\sqrt{x^{2}+1}+2x}=\lim\limits_{x\to -\infty}\frac{\left( x-1 \right)\left| x \right|\sqrt{1+\sqrt{\frac{1}{x}}+\frac{1}{x^{2}}}}{\left| x \right|\sqrt{1+\frac{1}{x^{2}}}+2x}=\lim\limits_{x\to -\infty}\frac{-\left( x-1 \right)\sqrt{1+\sqrt{\frac{1}{x}}+\frac{1}{x^{2}}}}{-\sqrt{1+\frac{1}{x^{2}}}+2}=+\infty$
d. $\lim\limits_{x\to +\infty}x^{2}\left( \sqrt[3]{x^{3}+1}-x \right)=\lim\limits_{x\to +\infty}\frac{x^{2}\left( \sqrt[3]{x^{3}+1}-x \right)\left( \left( \sqrt[3]{x^{3}+1} \right)^{2}+x.\sqrt[3]{x^{3}+1}+x^{2} \right)}{\left( \left( \sqrt[3]{x^{3}+1} \right)^{2}+x.\sqrt[3]{x^{3}+1}+x^{2} \right)}$
$=\lim\limits_{x\to +\infty}\frac{x^{2}}{\left( \left( \sqrt[3]{x^{3}+1} \right)^{2}+x.\sqrt[3]{x^{3}+1}+x^{2} \right)}=\lim\limits_{x\to +\infty}\frac{1}{\left( \sqrt[3]{1+\frac{1}{x^{3}}} \right)^{2}+\sqrt[3]{1+\frac{1}{x^{3}}}+1}=\frac{1}{3}$
b. $\lim\limits_{x\to -\infty}\frac{\left( x-1 \right)\left( 2x-1 \right)\left( 3x-1 \right)\left( 4x-1 \right)\left( 5x-1 \right)}{\left( 4x+5 \right)^{5}}=\lim\limits_{x\to -\infty}\frac{\left( 1-\frac{1}{x} \right)\left( 2-\frac{1}{x} \right)\left( 3-\frac{1}{x} \right)\left( 4-\frac{1}{x} \right)\left( 5-\frac{1}{x} \right)}{\left( 4+\frac{5}{x} \right)^{5}}=\frac{15}{128}$
c. $\lim\limits_{x\to -\infty}\frac{\left( x-1 \right)\sqrt{x^{2}+x\sqrt{x}+1}}{\sqrt{x^{2}+1}+2x}=\lim\limits_{x\to -\infty}\frac{\left( x-1 \right)\left| x \right|\sqrt{1+\sqrt{\frac{1}{x}}+\frac{1}{x^{2}}}}{\left| x \right|\sqrt{1+\frac{1}{x^{2}}}+2x}=\lim\limits_{x\to -\infty}\frac{-\left( x-1 \right)\sqrt{1+\sqrt{\frac{1}{x}}+\frac{1}{x^{2}}}}{-\sqrt{1+\frac{1}{x^{2}}}+2}=+\infty$
d. $\lim\limits_{x\to +\infty}x^{2}\left( \sqrt[3]{x^{3}+1}-x \right)=\lim\limits_{x\to +\infty}\frac{x^{2}\left( \sqrt[3]{x^{3}+1}-x \right)\left( \left( \sqrt[3]{x^{3}+1} \right)^{2}+x.\sqrt[3]{x^{3}+1}+x^{2} \right)}{\left( \left( \sqrt[3]{x^{3}+1} \right)^{2}+x.\sqrt[3]{x^{3}+1}+x^{2} \right)}$
$=\lim\limits_{x\to +\infty}\frac{x^{2}}{\left( \left( \sqrt[3]{x^{3}+1} \right)^{2}+x.\sqrt[3]{x^{3}+1}+x^{2} \right)}=\lim\limits_{x\to +\infty}\frac{1}{\left( \sqrt[3]{1+\frac{1}{x^{3}}} \right)^{2}+\sqrt[3]{1+\frac{1}{x^{3}}}+1}=\frac{1}{3}$