\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{4x+5}+x}{x^2+3x+2}=\dfrac{\sqrt{4.1+5}+1}{1^2+3.1+2}=\dfrac{2}{3}\)
(Đề là \(x\rightarrow-1\) thì hợp lý hơn)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{4x+5}+x}{x^2+3x+2}=\lim\limits_{x\rightarrow-1}\dfrac{\left(\sqrt{4x+5}+x\right)\left(\sqrt{4x+5}-x\right)}{\left(x+1\right)\left(x+2\right)\left(\sqrt{4x+5}-x\right)}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{-x^2+4x+5}{\left(x+1\right)\left(x+2\right)\left(\sqrt{4x+5}-x\right)}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(5-x\right)}{\left(x+1\right)\left(x+2\right)\left(\sqrt{4x+5}-x\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{5-x}{\left(x+2\right)\left(\sqrt{4x+5}-x\right)}=\dfrac{6}{1\left(1+1\right)}=3\)
