Question Number 27787 by abdo imad last updated on 14/Jan/18
$${find}\:\:{lim}_{{x}−>\mathrm{0}} \frac{{e}^{{x}} \:\:−{x}−\mathrm{1}}{{x}^{\mathrm{2}} }\:. \\ $$
Answered by prakash jain last updated on 14/Jan/18
$${e}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+… \\ $$$$\frac{{e}^{{x}} −{x}−\mathrm{1}}{{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}!}+\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{i}} }{\left({i}+\mathrm{2}\right)!} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(..\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$