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Here is the function:

$$ y = b^{b^{-2}} $$

It seems $\lim _{b \rightarrow \infty} y = 1$, but how do you prove this?

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  • $\begingroup$ When the base and exponent varies a trick is take logarithm $\endgroup$ – Valerin Dec 8 '13 at 14:24
  • $\begingroup$ @qed i have fixed my error,right it is approaching to one after taking logarithm of both side and calculate limit of right side equation using well known L'Hôpital's rule $\endgroup$ – dato datuashvili Dec 8 '13 at 14:32
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You can write $$ y=b^{b^{-2}}=e^{b^{-2}\log b} $$ and observing that, using l'Hôpital's rule, $$\lim_{b\to\infty}\frac{\log b}{b^2}=\lim_{b\to\infty}\frac{1}{2b^2}=0$$ and finally $$\lim_{b\to\infty}b^{b^{-2}}=e^0=1.$$

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  • $\begingroup$ i have fixed my error $\endgroup$ – dato datuashvili Dec 8 '13 at 14:30
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Taking log both sides you have :

$$ \ln y = \lim_{b\rightarrow \infty}\frac{\ln b}{b^2}$$

We know that $\ln b$ grows much much slowly compared to that of $b^2$ , so as $b$ approaches $\infty$ it the RHS approaches 0 and hence $\ln y = 0 $ and $y =1$

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aa i understand ,,then it is very simple ,ones again take log on both side ,we get

$log(y)=log(b)/(b^2)$,because second part is infinity over infinity ,take L'Hôpital's rule,but i think that is it approaching to $0$,not $1$,please check carrefuly,you are geting $y=e/b^2$,sorry i have made mistake

right $ln(y)=0$ and $y=1$,sorry i forgot that $y$ was in $ln$ function

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