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

Answer 1

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#

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Answer 2

Paisley Johnson

#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#

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Answer 3

William Abdalla

To find the limit of ln(x)/(x-1) as x approaches 1, we can use L'Hôpital's Rule. Taking the derivative of both the numerator and denominator, we get (1/x)/(1) = 1/x. Substituting x=1 into this expression, we find that the limit is equal to 1. Therefore, the limit of ln(x)/(x-1) as x approaches 1 is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit #lnx/(x-1)# as #x-&gt;1#?

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