Question

tính các giới hạn

a/lim\(\frac{\sqrt{x}-1}{x-1}\) (x-->1)

b/\(\frac{\sqrt{4x+1}-3}{x^2-4}\) (x-->2)

giúp mình 2 câu này vs ạ

Answer 1

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

\(\lim\limits_{x\rightarrow2}\frac{\sqrt{4x+1}-3}{x^2-4}=\lim\limits_{x\rightarrow2}\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\lim\limits_{x\rightarrow2}\frac{4}{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\frac{1}{6}\)