Question Number 78623 by mathmax by abdo last updated on 19/Jan/20
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\sqrt{\mathrm{1}+{x}+{x}^{\mathrm{2}} +….+{x}^{{n}} }\:\:−\mathrm{1}}{{x}^{\frac{{n}}{\mathrm{2}}} } \\ $$
Answered by jagoll last updated on 19/Jan/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +…+\mathrm{x}^{\mathrm{n}} \right)−\mathrm{1}}{\left(\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} +…+\mathrm{x}^{\mathrm{n}} \:}\:+\mathrm{1}\right)\:\mathrm{x}^{\frac{\mathrm{n}}{\mathrm{2}}} }\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{1}−\frac{\mathrm{n}}{\mathrm{2}}} +\mathrm{x}^{\mathrm{2}−\frac{\mathrm{n}}{\mathrm{2}}} +\mathrm{x}^{\mathrm{3}−\frac{\mathrm{n}}{\mathrm{2}}} +…+\mathrm{x}^{\frac{\mathrm{n}}{\mathrm{2}}} }{\left(\mathrm{2}\:\right)}\:=\:\mathrm{0} \\ $$$$ \\ $$