Question

Find limit using L'Hopital Rule
lim x-> a x - a / ln x - ln a

Answer 1

hmm why not write
\[\lim_{x\to a}\frac{\ln(x)-\ln(a)}{x-a}\]

Answer 2

is it the same? i dk

Answer 3

well no it is not the same, it is the reciprocal

Answer 4

i was just trying to point out that you don't need l'hopital, because this is the definition of the derivative of the log at \(x=a\)

Answer 5

umm..not sure about the definition of log..didnt memorize any of those

Answer 6

the derivative of \(\ln(x)\) is \(\frac{1}{x}\) replace \(x\) by \(a\) get \(\frac{1}{a}\) reciprocal is \(a\)

Answer 7

it is not about the definition of the log, it is about the definition of the derivative
\[f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\]

Answer 8

oh

Answer 9

but in any case you can use l'hopital if you like
the derivative of \(x-a\) is \(1\) and the derivative of \(\ln(x)-\ln(a)\) is \(\frac{1}{x}\)
giving you
\[\lim_{x\to a}\frac{1}{\frac{1}{x}}=\lim_{x\to a}x=a\]

Answer 10

will ln x = ln a in this case?

Answer 11

oh nevermind, a is a constant i think?