limitsssss

hereL = \lim_{x\rightarrow 0^{+}}e^{log([logcotx]^{tanx})}

now
\lim_{x\rightarrow 0}\frac{log(logcotx)}{cotx }=\lim_{cotx\rightarrow infty}\frac{log(logcotx)}{cotx }

since as x→0⁺, cotx →∞
=\lim_{cotx\rightarrow infty}\frac{log(logcotx)}{logcotx }\times \frac{logcotx}{cotx}

=0 \times 0 = 0

since
\lim_{x\rightarrow infty}\frac{logx}{x} = 0

so L = e⁰ = 1