Question

Tìm \(\lim\limits_{x\to -∞} f(x)=\dfrac{\sqrt{x^2+3x+5}}{4x-1}\) 

Answer 1

\(\lim\limits_{x\rightarrow-\infty}f\left(x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2+3x+5}}{4x-1}\)

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

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\cdot\sqrt{1+\dfrac{3}{x}+\dfrac{5}{x^2}}}{x\left(4-\dfrac{1}{x}\right)}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{1+\dfrac{3}{x}+\dfrac{5}{x^2}}}{4-\dfrac{1}{x}}\)

\(=\dfrac{-\sqrt{1+0+0}}{4-0}=\dfrac{-1}{4}\)