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#