Question

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

Answer 1

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

Answer 2

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

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

Answer 3

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