Bài tập nâng cao · Bài 20
Bài tập nâng cao Giới hạn của hàm số · Bài 20
Tìm các giới hạn sau:
a. $\lim\limits_{x\to 0}\frac{\sqrt[3]{1+x}-\sqrt{1-x}}{x}$
b. $\lim\limits_{x\to 0}\frac{\sqrt{3x+4}-\sqrt[3]{8+5x}}{x}$
c. $\lim\limits_{x\to 2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^{2}-3x+2}$
d. $\lim\limits_{x\to 0}\frac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^{2}}.$
a. $\lim\limits_{x\to 0}\frac{\sqrt[3]{1+x}-\sqrt{1-x}}{x}$
b. $\lim\limits_{x\to 0}\frac{\sqrt{3x+4}-\sqrt[3]{8+5x}}{x}$
c. $\lim\limits_{x\to 2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^{2}-3x+2}$
d. $\lim\limits_{x\to 0}\frac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^{2}}.$
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Lời giải
a. $\lim\limits_{x\to 0}\frac{\sqrt[3]{1+x}-\sqrt{1-x}}{x}=\lim\limits_{x\to 0}\frac{\sqrt[3]{1+x}-1+1-\sqrt{1-x}}{x}=\frac{5}{6}.$
b. $\lim\limits_{x\to 0}\frac{\sqrt{3x+4}-\sqrt[3]{8+5x}}{x}=\lim\limits_{x\to 0}\frac{\sqrt{3x+4}-2+2-\sqrt[3]{8+5x}}{x}=\frac{1}{3}.$
c. $\lim\limits_{x\to 2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^{2}-3x+2}=\lim\limits_{x\to 2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left( x-1 \right)\left( x-2 \right)}.$
$=\lim\limits_{x\to 2}\left[ \frac{8}{\left( x-1 \right)\left( \sqrt[3]{\left( 8x+11 \right)^{2}}+3\sqrt[3]{8x+11}+9 \right)}-\frac{1}{\left( x-1 \right)\left( 3+\sqrt{x+7} \right)} \right]=\frac{8}{27}-\frac{1}{6}=\frac{7}{54}.$
d. $\lim\limits_{x\to 0}\frac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^{2}}=\lim\limits_{x\to 0}\frac{\sqrt{1+4x}-\left( 2x+1 \right)+\left( 2x+1 \right)-\sqrt[3]{1+6x}}{x^{2}}$
$=\lim\limits_{x\to 0}\left[ \frac{-4}{\left( \sqrt{1+4x}+\left( 2x+1 \right) \right)}+\frac{8x+12}{\left( \left( 2x+1 \right)^{2}+\left( 2x+1 \right)\sqrt[3]{1+6x}+\sqrt[3]{\left( 1+6x \right)^{2}} \right)} \right]=\frac{-4}{2}+\frac{12}{3}=2.$
b. $\lim\limits_{x\to 0}\frac{\sqrt{3x+4}-\sqrt[3]{8+5x}}{x}=\lim\limits_{x\to 0}\frac{\sqrt{3x+4}-2+2-\sqrt[3]{8+5x}}{x}=\frac{1}{3}.$
c. $\lim\limits_{x\to 2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^{2}-3x+2}=\lim\limits_{x\to 2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left( x-1 \right)\left( x-2 \right)}.$
$=\lim\limits_{x\to 2}\left[ \frac{8}{\left( x-1 \right)\left( \sqrt[3]{\left( 8x+11 \right)^{2}}+3\sqrt[3]{8x+11}+9 \right)}-\frac{1}{\left( x-1 \right)\left( 3+\sqrt{x+7} \right)} \right]=\frac{8}{27}-\frac{1}{6}=\frac{7}{54}.$
d. $\lim\limits_{x\to 0}\frac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^{2}}=\lim\limits_{x\to 0}\frac{\sqrt{1+4x}-\left( 2x+1 \right)+\left( 2x+1 \right)-\sqrt[3]{1+6x}}{x^{2}}$
$=\lim\limits_{x\to 0}\left[ \frac{-4}{\left( \sqrt{1+4x}+\left( 2x+1 \right) \right)}+\frac{8x+12}{\left( \left( 2x+1 \right)^{2}+\left( 2x+1 \right)\sqrt[3]{1+6x}+\sqrt[3]{\left( 1+6x \right)^{2}} \right)} \right]=\frac{-4}{2}+\frac{12}{3}=2.$