Question Number 195840 by NANIGOPAL last updated on 11/Aug/23
$${find}\:{the}\:{valur} \\ $$$$\:\:{li}\underset{\rightarrow\mathrm{0}} {{m}}\:\frac{{x}−{sinx}}{{x}^{\mathrm{3}} } \\ $$
Commented by kokeb last updated on 11/Aug/23
$${find}\:{the}\:{valur} \\ $$$$\:\:{li}\underset{\rightarrow\mathrm{0}} {{m}}\:\frac{{x}−{sinx}}{{x}^{\mathrm{3}} }={li}\underset{\rightarrow\mathrm{0}} {{m}}\frac{\mathrm{1}−{cosx}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$={li}\underset{\rightarrow\mathrm{0}} {{m}}\frac{{sinx}}{\mathrm{4}{x}}=\frac{\mathrm{1}}{\mathrm{4}}{li}\underset{\rightarrow\mathrm{0}} {{m}}\frac{{sinx}}{{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by jabarsing last updated on 11/Aug/23
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{3}} }\:\overset{{LH}} {=}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{3}{x}^{\mathrm{2}} }\overset{{LH}} {=}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}}{\mathrm{6}{x}}= \\ $$$$\overset{{LH}} {=}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:{x}}{\mathrm{6}}\:=\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by MM42 last updated on 11/Aug/23
$$\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by jabarsing last updated on 11/Aug/23
$${or}:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{sinx}}{{x}^{\mathrm{3}} }\:\:\:\overset{{M}.{S}} {=}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}+…\right)}{{x}^{\mathrm{3}} }= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\cancel{{x}}−\cancel{{x}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}−\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}−…}{{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}!}−\mathrm{0}+\mathrm{0}−\mathrm{0}+…=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$$$ \\ $$