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`lim_(x->1) (log_3 3x)^(log_x 3)=`A. eB. `(1)/(e)`C. `1`D. `-(1)/(e)` |
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Answer» Correct Answer - A We have, ` lim_(xto1) (log _3 3x)^log) x3` ` =lim_(xto1)(log _(3)3+log_(3)x)^log)x3` `lim_(xto1) (1+log_(3)x)(1)/(log_3x)=e ^(lim_(xto1))log _(3xx -(1)/(log_(3)x)=e^1=e`. |
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