Bài tập nâng cao · Bài 4
Bài tập nâng cao Giới hạn của hàm số · Bài 4
Tính giới hạn
a. $\lim\limits_{x\to 2}\frac{\sqrt{2x+5}-3}{\sqrt{x+2}-2}$
b. $\lim\limits_{x\to 1}\frac{x^{3}-\sqrt{3x-2}}{x-1}$
a. $\lim\limits_{x\to 2}\frac{\sqrt{2x+5}-3}{\sqrt{x+2}-2}$
b. $\lim\limits_{x\to 1}\frac{x^{3}-\sqrt{3x-2}}{x-1}$
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Lời giải
a. $\lim\limits_{x\to 2}\frac{\sqrt{2x+5}-3}{\sqrt{x+2}-2}=\lim\limits_{x\to 2}\frac{\left( \sqrt{2x+5}-3 \right)\left( \sqrt{2x+5}+3 \right)\left( \sqrt{x+2}+2 \right)}{\left( \sqrt{x+2}-2 \right)\left( \sqrt{2x+5}+3 \right)\left( \sqrt{x+2}+2 \right)}$
$=\lim\limits_{x\to 2}\frac{\left( 2x+5-9 \right)\left( \sqrt{x+2}+2 \right)}{\left( x+2-4 \right)\left( \sqrt{2x+5}+3 \right)}=\lim\limits_{x\to 2}\frac{2\left( \sqrt{x+2}+2 \right)}{\sqrt{2x+5}+3}=\frac{4}{3}$
b. $\lim\limits_{x\to 1}\frac{x^{3}-\sqrt{3x-2}}{x-1}=\lim\limits_{x\to 1}\frac{\left( x^{3}-1 \right)-\left( \sqrt{3x-2}-1 \right)}{x-1}=\lim\limits_{x\to 1}\left[ \frac{x^{3}-1}{x-1}-\frac{\sqrt{3x-2}-1}{x-1} \right]$
$=\lim\limits_{x\to 1}\left[ x^{2}+x+1-\frac{3x-2-1}{\left( x-1 \right)\left( \sqrt{3x-2}+1 \right)} \right]=\lim\limits_{x\to 1}\left[ x^{2}+x+1-\frac{3}{\sqrt{3x-2}+1} \right]=3-\frac{3}{2}=\frac{3}{2}$
$=\lim\limits_{x\to 2}\frac{\left( 2x+5-9 \right)\left( \sqrt{x+2}+2 \right)}{\left( x+2-4 \right)\left( \sqrt{2x+5}+3 \right)}=\lim\limits_{x\to 2}\frac{2\left( \sqrt{x+2}+2 \right)}{\sqrt{2x+5}+3}=\frac{4}{3}$
b. $\lim\limits_{x\to 1}\frac{x^{3}-\sqrt{3x-2}}{x-1}=\lim\limits_{x\to 1}\frac{\left( x^{3}-1 \right)-\left( \sqrt{3x-2}-1 \right)}{x-1}=\lim\limits_{x\to 1}\left[ \frac{x^{3}-1}{x-1}-\frac{\sqrt{3x-2}-1}{x-1} \right]$
$=\lim\limits_{x\to 1}\left[ x^{2}+x+1-\frac{3x-2-1}{\left( x-1 \right)\left( \sqrt{3x-2}+1 \right)} \right]=\lim\limits_{x\to 1}\left[ x^{2}+x+1-\frac{3}{\sqrt{3x-2}+1} \right]=3-\frac{3}{2}=\frac{3}{2}$