Question Number 120068 by bramlexs22 last updated on 29/Oct/20
$$\left({i}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}{{x}^{\mathrm{4}} }\: \\ $$$$\left({ii}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −\mathrm{1}−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{{x}^{\mathrm{4}} } \\ $$$$\left({iii}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}−{x}}{\mathrm{arc}\:\mathrm{sin}\:{x}−{x}} \\ $$
Answered by Dwaipayan Shikari last updated on 29/Oct/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{24}}−\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}{{x}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{24}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cosx}=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!} \\ $$
Answered by Dwaipayan Shikari last updated on 29/Oct/20
$$\frac{{e}^{{x}} −\mathrm{1}−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{{x}^{\mathrm{4}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+\frac{{x}^{\mathrm{4}} }{\mathrm{24}}−\mathrm{1}−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{{x}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}} \\ $$