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

$$\mathit{E}valuate$$$$\underset{x\to 0^{+}}{\lim }{x}^m{\left(logx\right)}^n,m,n\in \mathbb{N}$$

Answered by MJS last updated on 07/May/18

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

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

$$Isthismethodcorrect?$$$$YouareapplyingL-hopitalat$$$$\frac{\ln x}{x^{\frac{-m}{n}}}buttheoriginalexpressionto$$$$whichweshouldapplyis$$$${\left(\frac{lnx}{x^{\frac{-m}{n}}}\right)}^n$$$$!!!!$$

Commented by MJS last updated on 07/May/18

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

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

$$Abdo,Tanmaysir,cananyonepls$$$$confirmtheabovestatement.$$$$(Isitalwaystrue?)$$$$Ihaveheardthat:$$$$\underset{x\to 0}{\lim }f\left(g\left(x\right)\right)=f\left(\underset{x\to 0}{\lim }g\left(x\right)\right)onlywhen$$$$fiscontinousatthatpoint.$$

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

$$yessirrahulfmustbecontinuetoapllythis$$$$result...$$

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