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limx->0 (cot x)1/logx  ?

ans: 0(m getting 1/e)

in Calculus by Active (2.6k points) | 1.1k views

3 Answers

0 votes
Alright ! Getting 1/e only. Where have u seen the answer from ?
by (293 points)
0
from b.s grewal book
0 votes
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
by (249 points)
+1
log y = -1 will give y = 1/e not 0.
0
yes i am also getting (1/e) as answer here
0 votes
$\lim_{x \to 0} \cot(x)^{\frac{1}{\ln(x)}} = \lim_{x \to 0} e^{\frac{1}{\ln(x)}\times \ln(\cot(x)) }\\ \lim_{x \to 0}\frac{\ln(\cot(x))}{\ln(x)} = -1\\ \lim_{u \to -1}e^u = e^-1 = \frac{1}{e}$
by Junior (985 points)
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