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Question-192033




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

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