Question Number 222427 by Nadirhashim last updated on 26/Jun/25
$$\:\:\boldsymbol{{if}}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{x}}^{\mathrm{3}} }+\frac{\boldsymbol{{a}}}{\boldsymbol{{x}}^{\mathrm{2}} }+\boldsymbol{{b}}\right)=\mathrm{1}\: \\ $$$$\:\:\:\:\boldsymbol{{find}}\:\boldsymbol{{a}}\:\boldsymbol{{and}}\:\boldsymbol{{b}}\:\:\boldsymbol{{without}} \\ $$$$\:\:\:\:\:\:\boldsymbol{{using}}\:\boldsymbol{{LH}}{opial}\:{rule} \\ $$
Answered by mr W last updated on 26/Jun/25
$$\mathrm{sin}\:{x}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}+… \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{2}{x}}{{x}^{\mathrm{3}} }+\frac{{a}}{{x}^{\mathrm{2}} }+{b}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}{x}−\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{32}{x}^{\mathrm{5}} }{\mathrm{5}!}−…}{{x}^{\mathrm{3}} }+\frac{{a}}{{x}^{\mathrm{2}} }+{b}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({b}+\frac{\mathrm{2}+{a}}{{x}^{\mathrm{2}} }−\frac{\mathrm{8}}{\mathrm{3}!}+\frac{\mathrm{32}{x}^{\mathrm{2}} }{\mathrm{5}!}−…\right) \\ $$$$={b}−\frac{\mathrm{8}}{\mathrm{3}!}+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}+{a}}{{x}^{\mathrm{2}} }\right)\overset{!} {=}\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}+{a}=\mathrm{0}\:\Rightarrow{a}=−\mathrm{2} \\ $$$$\Rightarrow{b}−\frac{\mathrm{8}}{\mathrm{3}!}=\mathrm{1}\:\Rightarrow{b}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$