How do you find the limit of $lnt/sqrt(1-t)$ as $t->1^-$ ?

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Answer 1 · 2017-01-12T15:40:05

$lim_(x->1^-) lnt/sqrt(1-t) = 0$

We have that:

$lim_(x->1^-) lnt = 0$

$lim_(x->1^-) sqrt(1-t) = 0$

so that the limit:

$lim_(x->1^-) lnt/sqrt(1-t)$

is in the indeterminate form $0/0$

We can then use l'Hospital's rule:

$lim_(x->1^-) lnt/sqrt(1-t) = lim_(x->1^-) (d/(dt)lnt)/(d/(dt)sqrt(1-t))= lim_(x->1^-) (1/t)/(-1/2 1/sqrt(1-t))=lim_(x->1^-) -2sqrt(1-t)/t=0$

How do you find the limit of lnt/sqrt(1-t)lnt/sqrt(1-t) as t->1^-t->1^-?