Question

[títtt]

tính giới hạn

a) \(\lim\limits_{x\rightarrow1}\dfrac{x^2-1}{\sqrt{3x+1}-2}\)

b)  \(\lim\limits_{x\rightarrow2}\dfrac{x^2-2x}{\sqrt{x+2}-2}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\lim\limits_{x\rightarrow1}\dfrac{x^2-1}{\sqrt{3x+1}-2}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)}{\dfrac{3x+1-4}{\sqrt{3x+1}+2}}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)\cdot\left(\sqrt{3x+1}+2\right)}{3\left(x-1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x+1\right)\left(\sqrt{3x+1}+2\right)}{3}\)

\(=\dfrac{\left(1+1\right)\left(\sqrt{3+1}+2\right)}{2}=\dfrac{2\cdot4}{3}=\dfrac{8}{3}\)

b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-2x}{\sqrt{x+2}-2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{\dfrac{x+2-4}{\sqrt{x+2}+2}}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)\cdot\left(\sqrt{x+2}+2\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2}x\left(\sqrt{x+2}+2\right)\)

\(=2\cdot\left(\sqrt{2+2}+2\right)\)

\(=2\cdot4=8\)