Question Number 34464 by rahul 19 last updated on 06/May/18
$$\boldsymbol{{E}}{valuate}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{x}^{{m}\:} \left({log}\:{x}\:\right)^{{n}} \:,\:{m},{n}\:\in\:\mathbb{N} \\ $$
Answered by MJS last updated on 07/May/18
$${x}^{{m}} \left(\mathrm{ln}\:{x}\right)^{{n}} =\left({x}^{\frac{\mathrm{m}}{\mathrm{n}}} \mathrm{ln}\:{x}\right)^{{n}} =\left(\frac{\mathrm{ln}\:{x}}{{x}^{−\frac{{m}}{{n}}} }\right)^{{n}} \\ $$$$\mathrm{l}'\mathrm{Hoptial}: \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{ln}\:{x}}{{x}^{−\frac{{m}}{{n}}} }=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\frac{\mathrm{1}}{{x}}}{−\frac{{m}}{{n}}{x}^{−\frac{{m}}{{n}}−\mathrm{1}} }=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}−\frac{{n}}{{m}}{x}^{\frac{{m}}{{n}}} =\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{x}^{{m}} \left(\mathrm{ln}\:{x}\right)^{{n}} =\mathrm{0} \\ $$
Commented by rahul 19 last updated on 07/May/18
$${Is}\:{this}\:\:{method}\:{correct}\:? \\ $$$${You}\:{are}\:{applying}\:{L}−{hopital}\:{at} \\ $$$$\frac{\mathrm{ln}\:{x}}{{x}^{\frac{−{m}}{{n}}} }\:{but}\:{the}\:{original}\:{expression}\:{to} \\ $$$${which}\:{we}\:{should}\:{apply}\:{is} \\ $$$$\left(\frac{{ln}\:{x}}{{x}^{\frac{−{m}}{{n}}} }\right)^{{n}} \: \\ $$$$!!!! \\ $$
Commented by MJS last updated on 07/May/18
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct}\:\mathrm{because}\:\mathrm{if}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of} \\ $$$${f}\left({x}\right)\:\mathrm{is}\:{l}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of}\:\left({f}\left({x}\right)\right)^{{n}} \:\mathrm{with}\:{n}\in\mathbb{N}\:\mathrm{should} \\ $$$$\mathrm{be}\:{l}^{{n}} … \\ $$
Commented by rahul 19 last updated on 07/May/18
$${Abdo},\:{Tanmay}\:{sir},{can}\:{anyone}\:{pls} \\ $$$${confirm}\:{the}\:{above}\:{statement}\:. \\ $$$$\left({Is}\:{it}\:{always}\:{true}?\right) \\ $$$${I}\:{have}\:{heard}\:{that}\:: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({g}\left({x}\right)\right)={f}\left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{g}\left({x}\right)\right)\:{only}\:{when} \\ $$$${f}\:{is}\:{continous}\:{at}\:{that}\:{point}. \\ $$
Commented by math khazana by abdo last updated on 08/May/18
$${yes}\:{sir}\:{rahul}\:{f}\:{must}\:{be}\:{continue}\:{to}\:{aplly}\:{this}\: \\ $$$${result}… \\ $$