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`lim_(x->1) (log_3 3x)^(log_x 3)=`
A. e
B. `(1)/(e)`
C. `1`
D. `-(1)/(e)`

1 Answer

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Best 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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