NL

Tìm limx➝+∞ \(\dfrac{\sqrt{x^2+2x}+3x}{\sqrt{4x^2+1}-x+2}\)

NT
2 tháng 12 2023 lúc 19:55

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

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

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