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To find lim x --> 5 [ ln(x-4)/(x-5)], if we substitute x = 5 , we get ln 1/0 or 0/0 which is an indeterminate form. We can use l'Hopital's rule here and use the derivatives of the numerator and the denominator.
lim x--> 5 [ (1/(x - 4)) /1]
substitute x = 5
=> 1 / 1 = 1
The value of lim x --> 5 [ ln(x-4)/(x-5)] = 1
First, we'll substitute x by the value of accumulation point.
lim ln(x-4)/(x-5) = ln(5-4)/(5-5) = ln1/0 = 0/0
We've get an indetermination, so, we'll apply L'Hospital rule:
lim ln(x-4)/(x-5) = lim d[ln(x-4)]/dx/d(x-5)/dx
lim d[ln(x-4)]/dx/d(x-5)/dx = lim 1/(x-4)/1
We'll substitute again x by 5 and we'll get:
lim 1/(x-4)/1 = 1/(5-4) = 1
The good news is that the result provided is exact, therefore, when x approaches to 5, the lim ln(x-4)/(x-5) = 1.
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