Question

Tính giới hạn sau:

\(\lim\limits_{x\rightarrow1}\dfrac{\left(x^2+3x+1\right)\sqrt{1+3x}-10}{x^2-1}\)

Answer 1

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x^2+3x+1\right)\left(\sqrt{3x+1}-2\right)+2\left(x^2+3x+1\right)-10}{x^2-1}\)

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

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