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q: how can i calculate the limit for $y=left(frac{1}{2}+frac{x}{ln x}right)^{frac{ln^2 x}{4ln^2ln x}}, xrightarrow infty$ how can i calculate the limit for this sequence: $$y=left(frac{1}{2}+frac{x}{ln x}right)^{frac{ln^2 x}{4ln^2ln x}}, xrightarrow infty$$ i have tried to solve it with using stolz theorem and i have arrived to the fact that: $$lim_{xrightarrow infty} y = lim_{xrightarrow infty} left(frac{1}{2}+frac{x}{ln x}right)^{frac{ln^2 x}{4ln^2ln x}} = e^{e}$$ when using wolframalpha i have got $exp(e)$. but i have noticed that the answer also is $e^e$. is there something that i have missed? a: the function $frac{ln^2 x}{4ln^2ln x}$ is decreasing for $x > 1$. the logarithmic function $ln x$ is increasing for $x > 1$. therefore $$ln xleq lnln x$$ for $x > 1$. this means that $$frac{ln^2 x}{4ln^2ln x}geq frac{ln^2ln x}{4ln^2lnln x}=frac{lnln x}{2lnlnln x}.$$ now the function $f(x)=frac{lnln x}{2lnlnln x}$ is decreasing for $x > e$ and therefore $f(x)leq f(e)=frac{1}{2}$.

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