Bài tập nâng cao · Bài 22
Bài tập nâng cao Giới hạn của hàm số · Bài 22
Tìm giới hạn
a. $\lim\limits_{x\to -\infty}\frac{\sqrt[3]{x^{6}-x^{4}+1}+\sqrt{x^{6}+1}}{2x-1}$
b. $\lim\limits_{x\to +\infty}\frac{x\sqrt{x}+1}{x^{2}+x+1}$
c. $\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}+2x}+2-x}{\sqrt{9x^{2}+1}-x+2}$
d. $\lim\limits_{x\to \pm\infty}\frac{\sqrt{x^{2}+2x+3}+4x+1}{\sqrt{4x^{2}+1}+2-x}$
a. $\lim\limits_{x\to -\infty}\frac{\sqrt[3]{x^{6}-x^{4}+1}+\sqrt{x^{6}+1}}{2x-1}$
b. $\lim\limits_{x\to +\infty}\frac{x\sqrt{x}+1}{x^{2}+x+1}$
c. $\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}+2x}+2-x}{\sqrt{9x^{2}+1}-x+2}$
d. $\lim\limits_{x\to \pm\infty}\frac{\sqrt{x^{2}+2x+3}+4x+1}{\sqrt{4x^{2}+1}+2-x}$
Xem lời giải
Lời giải
a. $\lim\limits_{x\to -\infty}\frac{\sqrt[3]{x^{6}-x^{4}+1}+\sqrt{x^{6}+1}}{2x-1}=\lim\limits_{x\to -\infty}\frac{x^{2}\sqrt[3]{1-\frac{1}{x^{2}}+\frac{1}{x^{6}}}+\left| x^{3} \right|\sqrt{1+\frac{1}{x^{6}}}}{x\left( 2-\frac{1}{x} \right)}$
$=\lim\limits_{x\to -\infty}\frac{x\sqrt[3]{1-\frac{1}{x^{2}}+\frac{1}{x^{6}}}-x^{2}\sqrt{1+\frac{1}{x^{6}}}}{\left( 2-\frac{1}{x} \right)}=-\infty$
b. $\lim\limits_{x\to +\infty}\frac{x\sqrt{x}+1}{x^{2}+x+1}=\lim\limits_{x\to +\infty}\frac{\sqrt{\frac{1}{x}}+\frac{1}{x^{2}}}{1+\frac{1}{x}+\frac{1}{x^{2}}}=0$
c. $\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}+2x}+2-x}{\sqrt{9x^{2}+1}-x+2}=\lim\limits_{x\to -\infty}\frac{\left| x \right|\sqrt{1+\frac{2}{x}}+x\left( \frac{2}{x}-1 \right)}{\left| x \right|\sqrt{9+\frac{1}{x^{2}}}+x\left( -1+\frac{2}{x} \right)}=\lim\limits_{x\to -\infty}\frac{-x\sqrt{1+\frac{2}{x}}+x\left( \frac{2}{x}-1 \right)}{-x\sqrt{9+\frac{1}{x^{2}}}+x\left( -1+\frac{2}{x} \right)}$
$\lim\limits_{x\to -\infty}\frac{\frac{2}{x}-1-\sqrt{1+\frac{2}{x}}}{-\sqrt{9+\frac{1}{x^{2}}}-1+\frac{2}{x}}=\frac{1}{2}$
d. $\lim\limits_{x\to \pm\infty}\frac{\sqrt{x^{2}+2x+3}+4x+1}{\sqrt{4x^{2}+1}+2-x}$
+ $\lim\limits_{x\to +\infty}\frac{\sqrt{x^{2}+2x+3}+4x+1}{\sqrt{4x^{2}+1}+2-x}=\lim\limits_{x\to +\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^{2}}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^{2}}}+\frac{2}{x}-1}=5$
+ $\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}+2x+3}+4x+1}{\sqrt{4x^{2}+1}+2-x}=\lim\limits_{x\to -\infty}\frac{-\sqrt{1+\frac{2}{x}+\frac{3}{x^{2}}}+4+\frac{1}{x}}{-\sqrt{4+\frac{1}{x^{2}}}+\frac{2}{x}-1}=-1$
$=\lim\limits_{x\to -\infty}\frac{x\sqrt[3]{1-\frac{1}{x^{2}}+\frac{1}{x^{6}}}-x^{2}\sqrt{1+\frac{1}{x^{6}}}}{\left( 2-\frac{1}{x} \right)}=-\infty$
b. $\lim\limits_{x\to +\infty}\frac{x\sqrt{x}+1}{x^{2}+x+1}=\lim\limits_{x\to +\infty}\frac{\sqrt{\frac{1}{x}}+\frac{1}{x^{2}}}{1+\frac{1}{x}+\frac{1}{x^{2}}}=0$
c. $\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}+2x}+2-x}{\sqrt{9x^{2}+1}-x+2}=\lim\limits_{x\to -\infty}\frac{\left| x \right|\sqrt{1+\frac{2}{x}}+x\left( \frac{2}{x}-1 \right)}{\left| x \right|\sqrt{9+\frac{1}{x^{2}}}+x\left( -1+\frac{2}{x} \right)}=\lim\limits_{x\to -\infty}\frac{-x\sqrt{1+\frac{2}{x}}+x\left( \frac{2}{x}-1 \right)}{-x\sqrt{9+\frac{1}{x^{2}}}+x\left( -1+\frac{2}{x} \right)}$
$\lim\limits_{x\to -\infty}\frac{\frac{2}{x}-1-\sqrt{1+\frac{2}{x}}}{-\sqrt{9+\frac{1}{x^{2}}}-1+\frac{2}{x}}=\frac{1}{2}$
d. $\lim\limits_{x\to \pm\infty}\frac{\sqrt{x^{2}+2x+3}+4x+1}{\sqrt{4x^{2}+1}+2-x}$
+ $\lim\limits_{x\to +\infty}\frac{\sqrt{x^{2}+2x+3}+4x+1}{\sqrt{4x^{2}+1}+2-x}=\lim\limits_{x\to +\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^{2}}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^{2}}}+\frac{2}{x}-1}=5$
+ $\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}+2x+3}+4x+1}{\sqrt{4x^{2}+1}+2-x}=\lim\limits_{x\to -\infty}\frac{-\sqrt{1+\frac{2}{x}+\frac{3}{x^{2}}}+4+\frac{1}{x}}{-\sqrt{4+\frac{1}{x^{2}}}+\frac{2}{x}-1}=-1$