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lim
_{x>0}
(cot x)
^{1/logx}
?
ans: 0(m getting 1/e)
asked
Jul 21, 2015
in
Calculus
by
Vivek sharma
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3
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Alright ! Getting 1/e only. Where have u seen the answer from ?
answered
Jul 21, 2015
by
saloni
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293
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from b.s grewal book
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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 lhospitals
=lt x>0 x/(sinx*cosx)
=1
y=0
answered
Jul 21, 2015
by
Sudeep Choudhary
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249
points)
comment
+1
log y = 1 will give y = 1/e not 0.
0
yes i am also getting (1/e) as answer here
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$\lim_{x \to 0} \cot(x)^{\frac{1}{\ln(x)}} = \lim_{x \to 0} e^{\frac{1}{\ln(x)}\times \ln(\cot(x)) }\\ \lim_{x \to 0}\frac{\ln(\cot(x))}{\ln(x)} = 1\\ \lim_{u \to 1}e^u = e^1 = \frac{1}{e}$
answered
Aug 5, 2015
by
Salman
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