求证lim x→∞ √x^2+1-√x^-2x+5 =1
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求证limx→∞√x^2+1-√x^-2x+5=1求证limx→∞√x^2+1-√x^-2x+5=1求证limx→∞√x^2+1-√x^-2x+5=1分子有理化得原式=lim(x→∞)[(x^2+1)
求证lim x→∞ √x^2+1-√x^-2x+5 =1
求证lim x→∞ √x^2+1-√x^-2x+5 =1
求证lim x→∞ √x^2+1-√x^-2x+5 =1
分子有理化得
原式=lim(x→∞) [(x^2+1)-(x^2-2x+5)]/[√(x^2+1)+√(x^2-2x+5)]
=lim(x→∞) (2x+4)/[√(x^2+1)+√(x^2-2x+5)]
=lim(x→∞) (2x+4)/[√(x^2)+√(x^2)]
=1
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