Question Number 212137 by MrGaster last updated on 03/Oct/24
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt[{{n}}]{{n}}−\mathrm{1}\right)\sqrt{{n}}=? \\ $$$$ \\ $$
Answered by Frix last updated on 03/Oct/24
![lim_(n→∞) (n^((1/n)+(1/2)) −n^(1/2) ) =lim_(t→0^+ ) ((1−t^t )/t^(t+(1/2)) ) =_([& transform]) ^([l′Ho^ pital]) =−2lim_(t→0^+ ) ((t^(1/2) (1+ln t))/(1+2t(1+ln t))) =^((∗)) (0/1)=0 (∗) lim_(t→0) t^r ln t =0 ∀r>0](Q212138.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left({n}^{\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}}} −{n}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)\:=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{1}−{t}^{{t}} }{{t}^{{t}+\frac{\mathrm{1}}{\mathrm{2}}} }\:\underset{\left[\&\:\mathrm{transform}\right]} {\overset{\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right]} {=}} \\ $$$$=−\mathrm{2}\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+\mathrm{ln}\:{t}\right)}{\mathrm{1}+\mathrm{2}{t}\left(\mathrm{1}+\mathrm{ln}\:{t}\right)}\:\overset{\left(\ast\right)} {=}\frac{\mathrm{0}}{\mathrm{1}}=\mathrm{0} \\ $$$$ \\ $$$$\left(\ast\right)\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{t}^{{r}} \mathrm{ln}\:{t}\:=\mathrm{0}\:\forall{r}>\mathrm{0} \\ $$