Question Number 195029 by cortano12 last updated on 22/Jul/23
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Answered by MM42 last updated on 22/Jul/23
$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\frac{\sqrt{\mathrm{1}+{x}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}{ln}\left(\mathrm{1}+{x}\right)}\:=\overset{{hop}} {\rightarrow}\: \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+{x}}}}{\:\frac{\sqrt{\mathrm{3}}}{\mathrm{1}+{x}}}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\: \\ $$