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Question Number 34464 by rahul 19 last updated on 06/May/18

Evaluate   lim_(x→0^+ )  x^(m ) (log x )^n  , m,n ∈ N

$$\boldsymbol{{E}}{valuate}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{x}^{{m}\:} \left({log}\:{x}\:\right)^{{n}} \:,\:{m},{n}\:\in\:\mathbb{N} \\ $$

Answered by MJS last updated on 07/May/18

x^m (ln x)^n =(x^(m/n) ln x)^n =(((ln x)/x^(−(m/n)) ))^n   l′Hoptial:  lim_(x→0^+ ) ((ln x)/x^(−(m/n)) )=lim_(x→0^+ ) ((1/x)/(−(m/n)x^(−(m/n)−1) ))=lim_(x→0^+ ) −(n/m)x^(m/n) =0 ⇒  ⇒ lim_(x→0^+ ) x^m (ln x)^n =0

$${x}^{{m}} \left(\mathrm{ln}\:{x}\right)^{{n}} =\left({x}^{\frac{\mathrm{m}}{\mathrm{n}}} \mathrm{ln}\:{x}\right)^{{n}} =\left(\frac{\mathrm{ln}\:{x}}{{x}^{−\frac{{m}}{{n}}} }\right)^{{n}} \\ $$$$\mathrm{l}'\mathrm{Hoptial}: \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{ln}\:{x}}{{x}^{−\frac{{m}}{{n}}} }=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\frac{\mathrm{1}}{{x}}}{−\frac{{m}}{{n}}{x}^{−\frac{{m}}{{n}}−\mathrm{1}} }=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}−\frac{{n}}{{m}}{x}^{\frac{{m}}{{n}}} =\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{x}^{{m}} \left(\mathrm{ln}\:{x}\right)^{{n}} =\mathrm{0} \\ $$

Commented by rahul 19 last updated on 07/May/18

Is this  method correct ?  You are applying L−hopital at  ((ln x)/x^((−m)/n) ) but the original expression to  which we should apply is  (((ln x)/x^((−m)/n) ))^n    !!!!

$${Is}\:{this}\:\:{method}\:{correct}\:? \\ $$$${You}\:{are}\:{applying}\:{L}−{hopital}\:{at} \\ $$$$\frac{\mathrm{ln}\:{x}}{{x}^{\frac{−{m}}{{n}}} }\:{but}\:{the}\:{original}\:{expression}\:{to} \\ $$$${which}\:{we}\:{should}\:{apply}\:{is} \\ $$$$\left(\frac{{ln}\:{x}}{{x}^{\frac{−{m}}{{n}}} }\right)^{{n}} \: \\ $$$$!!!! \\ $$

Commented by MJS last updated on 07/May/18

I think it′s correct because if the limit of  f(x) is l the limit of (f(x))^n  with n∈N should  be l^n ...

$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct}\:\mathrm{because}\:\mathrm{if}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of} \\ $$$${f}\left({x}\right)\:\mathrm{is}\:{l}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of}\:\left({f}\left({x}\right)\right)^{{n}} \:\mathrm{with}\:{n}\in\mathbb{N}\:\mathrm{should} \\ $$$$\mathrm{be}\:{l}^{{n}} ... \\ $$

Commented by rahul 19 last updated on 07/May/18

Abdo, Tanmay sir,can anyone pls  confirm the above statement .  (Is it always true?)  I have heard that :  lim_(x→0) f(g(x))=f(lim_(x→0) g(x)) only when  f is continous at that point.

$${Abdo},\:{Tanmay}\:{sir},{can}\:{anyone}\:{pls} \\ $$$${confirm}\:{the}\:{above}\:{statement}\:. \\ $$$$\left({Is}\:{it}\:{always}\:{true}?\right) \\ $$$${I}\:{have}\:{heard}\:{that}\:: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({g}\left({x}\right)\right)={f}\left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{g}\left({x}\right)\right)\:{only}\:{when} \\ $$$${f}\:{is}\:{continous}\:{at}\:{that}\:{point}. \\ $$

Commented by math khazana by abdo last updated on 08/May/18

yes sir rahul f must be continue to aplly this   result...

$${yes}\:{sir}\:{rahul}\:{f}\:{must}\:{be}\:{continue}\:{to}\:{aplly}\:{this}\: \\ $$$${result}... \\ $$

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