1solve 2
1) \lim_{x\rightarrow 0} (\frac{sinx}{x}) ^{1/x}
2) \lim_{x\rightarrow +0} (log cotx ) ^{tanx}
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3 Answers
23hereL = \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