Question Number 192033 by sonukgindia last updated on 06/May/23
Answered by mehdee42 last updated on 06/May/23
$${L}={e}^{\:{lim}_{{x}\rightarrow\mathrm{0}} \:\left(\frac{{sinx}}{{x}}\:−\mathrm{1}\:\right)\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}} \:={e}^{\:{lim}_{{x}\rightarrow\mathrm{0}} \:\left(\frac{{sinx}\:−\mathrm{1}}{{x}^{\mathrm{3}} }\:\right)\:} =\:{e}^{\:{lim}_{{x}\rightarrow\mathrm{0}} \:\left(\frac{−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} }\right)} \:={e}^{−\frac{\mathrm{1}}{\mathrm{6}}} \\ $$
Answered by qaz last updated on 06/May/23
$$\mathrm{sin}\:{x}={x}−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +... \\ $$$$\Rightarrow{L}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} ={e}^{−\mathrm{1}/\mathrm{6}} \\ $$
Commented by mehdee42 last updated on 06/May/23
$${for}\:{the}\:{same}\:{reason}\:{of}\:{being}\:{vagio}\:\frac{{o}}{{o}}\:{or}\:\frac{\infty}{\infty} \\ $$
Commented by Subhi last updated on 06/May/23
$${why}\:{we}\:{can}'{t}\:{apply}\:{the}\:{limit}\:{for}\:{the}\:{base}\:{and}\:{the}\:{limit}\:{for}\:{the}\:{power}\: \\ $$$${that}\:{will}\:{give}\:\mathrm{1}^{\infty} \\ $$