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Alright ! Getting 1/e only. Where have u seen the answer from ?
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lt x->0 (cotx)^(1/logx)

lny=lt x->0 log(cotx)/(logx)

   =lt x->0 (-cosecx^2  * x/cotx) using l-hospitals

   =lt x->0 -x/(sinx*cosx)
   
   =-1
y=0
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$\lim_{x \to 0} \cot(x)^{\frac{1}{\ln(x)}} = \lim_{x \to 0} e^{\frac{1}{\ln(x)}\times \ln(\cot(x)) }\\ \lim_{x \to 0}\frac{\ln(\cot(x))}{\ln(x)} = -1\\ \lim_{u \to -1}e^u = e^-1 = \frac{1}{e}$

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