Bài tập nâng cao · Bài 5
Bài tập nâng cao Giới hạn của hàm số · Bài 5
Tính giới hạn
a. $\lim\limits_{x\to -1}\frac{\sqrt[4]{x+2}-1}{\sqrt[3]{x+2}-1}$
b. $\lim\limits_{x\to 1}\frac{\sqrt[3]{x+7}-\sqrt{x+3}}{x-1}$
a. $\lim\limits_{x\to -1}\frac{\sqrt[4]{x+2}-1}{\sqrt[3]{x+2}-1}$
b. $\lim\limits_{x\to 1}\frac{\sqrt[3]{x+7}-\sqrt{x+3}}{x-1}$
Xem lời giải
Lời giải
a. Đặt $t=\sqrt[12]{x+2}\Rightarrow x=t^{12}-2$ khi đó $x\to -1$ thì $t\to 1$. Do đó:
$\lim\limits_{x\to -1}\frac{\sqrt[4]{x+2}-1}{\sqrt[3]{x+2}-1}=\lim\limits_{t\to 1}\frac{t^{3}-1}{t^{4}-1}=\lim\limits_{t\to 1}\frac{\left( t-1 \right)\left( t^{2}+t+1 \right)}{\left( t-1 \right)\left( t+1 \right)\left( t^{2}+1 \right)}=\lim\limits_{t\to 1}\frac{t^{2}+t+1}{\left( t+1 \right)\left( t^{2}+1 \right)}=\frac{3}{4}$
b. $\lim\limits_{x\to 1}\frac{\sqrt[3]{x+7}-\sqrt{x+3}}{x-1}=\lim\limits_{x\to 1}\frac{\left( \sqrt[3]{x+7}-2 \right)-\left( \sqrt{x+3}-2 \right)}{x-1}=\lim\limits_{x\to 1}\left[ \frac{\sqrt[3]{x+7}-2}{x-1}-\frac{\sqrt{x+3}-2}{x-1} \right]$
$=\lim\limits_{x\to 1}\left[ \frac{x+7-2^{3}}{\left( x-1 \right)\left( \sqrt[3]{\left( x+7 \right)^{2}}+2\sqrt[3]{x+7}+4 \right)}-\frac{1}{\sqrt{x+3}+2} \right]$
$=\lim\limits_{x\to 1}\left[ \frac{1}{\sqrt[3]{\left( x+7 \right)^{2}}+2\sqrt[3]{x+7}+4}-\frac{1}{\sqrt{x+3}+2} \right]=\frac{1}{12}-\frac{1}{4}=-\frac{1}{6}$
$\lim\limits_{x\to -1}\frac{\sqrt[4]{x+2}-1}{\sqrt[3]{x+2}-1}=\lim\limits_{t\to 1}\frac{t^{3}-1}{t^{4}-1}=\lim\limits_{t\to 1}\frac{\left( t-1 \right)\left( t^{2}+t+1 \right)}{\left( t-1 \right)\left( t+1 \right)\left( t^{2}+1 \right)}=\lim\limits_{t\to 1}\frac{t^{2}+t+1}{\left( t+1 \right)\left( t^{2}+1 \right)}=\frac{3}{4}$
b. $\lim\limits_{x\to 1}\frac{\sqrt[3]{x+7}-\sqrt{x+3}}{x-1}=\lim\limits_{x\to 1}\frac{\left( \sqrt[3]{x+7}-2 \right)-\left( \sqrt{x+3}-2 \right)}{x-1}=\lim\limits_{x\to 1}\left[ \frac{\sqrt[3]{x+7}-2}{x-1}-\frac{\sqrt{x+3}-2}{x-1} \right]$
$=\lim\limits_{x\to 1}\left[ \frac{x+7-2^{3}}{\left( x-1 \right)\left( \sqrt[3]{\left( x+7 \right)^{2}}+2\sqrt[3]{x+7}+4 \right)}-\frac{1}{\sqrt{x+3}+2} \right]$
$=\lim\limits_{x\to 1}\left[ \frac{1}{\sqrt[3]{\left( x+7 \right)^{2}}+2\sqrt[3]{x+7}+4}-\frac{1}{\sqrt{x+3}+2} \right]=\frac{1}{12}-\frac{1}{4}=-\frac{1}{6}$