How do you find the limit $lnx/(x-1)$ as $x->1$ ?

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Answer 1 · 2016-10-19T17:47:34

The answer $=1$

Use l'Hôpital Rule
Limit $= 0/0$
So the derivative of numerator and denominator gives
$=(lnx')/((x-1)')=1/x$ and the limit is $1/1=1$ as $x->1$

Answer 2 · 2016-10-19T19:54:48

$1$

$lim_(x->0)log_e x/(x-1) = lim_(x->1)log_e(x^(1/(x-1)))$

Making $y = x-1$ we have $x = 1+y$ and $lim_(x->1) equiv lim_(y->0)$

then

$lim_(x->1)log_e(x^(1/(x-1)))equiv lim_(y->0)log_e((1+y)^(1/y)) = log_e e = 1$

How do you find the limit lnx/(x-1)lnx/(x-1) as x->1x->1?