Question

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

Answer 1

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

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

\(=\lim\limits_{x\rightarrow1}\dfrac{x+4}{\left(x+1\right)\left(\sqrt{x^2+3x}+2\right)}=\dfrac{5}{2\left(2+2\right)}=\dfrac{5}{8}\)