Evaluate `lim_(xto1) "sec" (pi)/(2^(x)).log_(e)x`.

1.	Evaluate `lim_(xto1) "sec" (pi)/(2^(x)).log_(e)x`.
Answer» We have, `underset(xto1)lim(log_(e)x)/("cos"(pi)/(2^(x)))=underset(xto1)lim(log(1+(x-1)))/(underset(xto1)lim(sin((pi)/(2)-(pi)/(2^(x))))/(((pi)/(2)-(pi)/(2^(x))))).underset(xto1)lim(x-1)/((pi)/(2)-(pi)/(2^(x)))` `=underset(xto1)lim(x-1)/(pi((2^(x-1)-1)/(2^(x))))` `=underset(xto1)lim(2^(x))/(pi).(x-1)/(2^(x-1)-1)` `=(2)/(pilog2)`

Discussion

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