How do you find #lim lnt/(t-1)# as #t->1# using l'Hospital's Rule?

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Answer 1 · 2017-02-16T16:06:03

The answer is #=1#

We use,

#(lnt)'=1/t#

#(t-1)'=1#

Normally,

#lim_(t->1)lnt/(t-1)=0/0#

This is undetermined

So, we apply L'Hospital's Rule

#lim_(t->1)lnt/(t-1)=lim_(t->1)((lnt)')/((t-1)')#

#=lim_(t->1)(1/t)/(1)=1#