1is the answer (a) ?????
For x>0, \lim_{x\rightarrow 0}\left[(sinx)^{\frac{1}{x}}+\left(\frac{1}{x} \right)^{sinx} \right] is
(a) 0
(b) -1
(c) 1
(d) 2
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4 Answers
1for x→0+ sinx→0+ and (1/x)→∞
so (sinx)(1/x) →0
please correct me. i dont know where i am going wrong
for (1/x)sinx taking logarithm and applying l'hospitals rule
ln L = lim sinx.ln(1/x)
ln L=lim -(lnx)/cosecx
applying lhspital rule
ln L= lim tanx(sinx/x)
ln L= .... tan x=0
i had made a mistake earlier here forgot that ln L=0 not L itself
so L=1