Question

Please, help! Need a hint:)
lim x-> a: (ln x - ln a) / (x - a)

Answer 1

lnx - lna = ln(x/a)

Answer 2

L hopsital to the rescue:
\[\lim_{x \rightarrow a} \frac{(\ln x - \ln a)}{ (x - a)} =>\lim_{x \rightarrow a} \frac{(\frac{1}{ x} )}{ (1)}\]
put x=a now :)

Answer 3

lol, using the derivative of ln(x) to define the derivative of ln(x) is not sound practice.

Answer 4

let x-a = h, then a = x-h

ln(x) - ln(x-h)
------------
h


ln(x/(x+h))
----------- ; doesnt seems at first glance a nicer rendition does it
h

Answer 5

can I do it without L'hopital? we are not supposed to know about it. Are there any other ways?

Answer 6

http://www.youtube.com/watch?v=Fk4C0L-5e3A

this walks you thru it pretty well by first principles

Answer 7

thank you, the video is good, but I'm still confused

Answer 8

on what point?

Answer 9

that x is approching not 0 or infinity but unknown a

Answer 10

do you know how to find the derivative of ln(x) using the definition of derivative?

Answer 11

let x = a+h and you will see that its the same setup

Answer 12

\[f'(x)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}\]

Answer 13

\[\lim_{x\to a}\frac{ln(x)-ln(a)}{x-a}\]
lets x=a+h

\[\lim_{a+h\to a}\frac{ln(a+h)-ln(a)}{a+h-a}\]

\[\lim_{h\to o}\frac{ln(a+h)-ln(a)}{h}\]

Answer 14

\[\lim_{h\to o}\frac{ln(a+h)-ln(a)}{h}\]
\[\lim_{h\to o}\frac1hln(a/a+h/a)\]
\[\lim_{h\to o}\frac1hln(1+\frac{h}{a})\]
they say, let v = h/a, h=va; as h to 0, v to 0
\[\lim_{v\to o}\frac1{va}ln(1+v)\]
\[\lim_{v\to o}\frac1{a}ln[(1+v)^{1/v}]\]
\[\frac1a~ln[~\lim_{v\to o}~(1+v)^{1/v}]\]

Answer 15

thank you,I finally understood why we do x = a+h (in order to have ->0)

Answer 16

:) youre welcome

Answer 17

if its prior knowledge that the inner ln limits to e, then we are set; other than that we would have to prolly prove that the iner ln does go to e :)

Answer 18

and at the end I just leave 1/a ln and count only the part in []?

Answer 19

at the end, you either already know that the inside of the ln[] goes to "e"; and that ln[e] = 1 ; or you would have to prove that it does go to e.

Answer 20

it goes to e because of the second principle?

Answer 21

in other words; if you already know by prior work that:\[\lim_{v\to0}(1+v)^{1/v}=e\]
your set, otherwise you would prolly have to prove this limits to e

Answer 22

I mean, it =1 because of 2nd principle

Answer 23

not sure what a second principle is in relation to this.

Answer 24

ln(e) = 1 by properties of logarithms

Answer 25

yeah, I got it finally, thank you!

Answer 26

oh good!, cause i was getting a little brain tied trying to proof it limited to e lol

Answer 27

i gotta get home now, so good luck and all

Answer 28

thank you! wish you all the best!!