Question Number 220852 by fantastic last updated on 20/May/25
$$\underset{{x}\rightarrow\mathrm{0}} {{Lim}}\left\{\frac{{xe}^{{x}} −{log}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{2}} }\right\} \\ $$
Answered by SdC355 last updated on 20/May/25
$$\frac{\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\left({xe}^{{x}} −\mathrm{ln}\left({x}+\mathrm{1}\right)\right)}{\frac{\mathrm{d}\:}{\mathrm{d}{x}}\:{x}^{\mathrm{2}} }=\frac{\left({x}+\mathrm{1}\right){e}^{{x}} −\frac{\mathrm{1}}{{x}+\mathrm{1}}}{\mathrm{2}{x}} \\ $$$$\frac{\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\left({x}+\mathrm{1}\right){e}^{{x}} −\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)}{\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\:\mathrm{2}{x}}=\frac{{e}^{{x}} +\left({x}+\mathrm{1}\right){e}^{{x}} +\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\therefore\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\left({x}+\mathrm{2}\right){e}^{{x}} +\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}. \\ $$
Commented by fantastic last updated on 20/May/25
$${right} \\ $$