Question

limit as x approaches 1+ ((1/x-1)-(1/lnx))

Answer 1

\[\lim_{x \rightarrow 1^+} (\frac{1}{x}-1-\frac{1}{\ln(x)}) \]

Answer 2

i think it is
\[\lim_{x\rightarrow 1^+}\frac{1}{x-1}-\frac{1}{\ln(x)}\]

Answer 3

but i could be wrong for sure

Answer 4

it is up to interpretation i believe
its like art

Answer 5

i know it is equal to -1/2, but I'm not sure of the steps.

Answer 6

would you like to tell us what the problem is?
does satellite have it right?

Answer 7

it would help if we knew it was what i wrote or what myininaya wrote, because they are different

Answer 8

sorry! what satellite wrote. damn typos...

Answer 9

i get
\[-\frac{1}{2}\]

Answer 10

start with
\[\frac{\ln(x)-x+1}{x\ln(x)-\ln(x)}\] then use l'hopital twice

Answer 11

\[\lim_{x \rightarrow 1^+}\frac{\ln(x)-(x-1)}{\ln(x) \cdot (x-1)}=\lim_{x \rightarrow 1^+}\frac{\ln(x)-x+1}{x \ln(x)-\ln(x)}\]
since if we plug in 1 we get 0/0 we can use lhospital
\[=\lim_{x \rightarrow 1}\frac{\frac{1}{x}-1}{\ln(x)+x \cdot \frac{1}{x}-\frac{1}{x}}\]
\[=\lim_{x \rightarrow 1}\frac{\frac{1}{x}-1}{\ln(x)-\frac{1}{x}+1}\]
we can do it again since we have 0/0 again
\[=\lim_{x \rightarrow 1}\frac{\frac{-1}{x^2}}{\frac{1}{x}-\frac{-1}{x^2}} =\lim_{x \rightarrow 1}\frac{-1}{x+1}\]

Answer 12

=-1/2

Answer 13

yikes

Answer 14

thanks so so much.

Answer 15

at step two feel free to multiply top and bottom by x to turn
\[=\lim_{x \rightarrow 1}\frac{\frac{1}{x}-1}{\ln(x)-\frac{1}{x}+1}\] into
\[\frac{1-x}{x\ln(x)+x-1}\] then use l'hopital again

Answer 16

why i don't have to use product rule with mine after that one step

Answer 17

myininaya usually loves to clear fractions and i do not. we traded places this time

Answer 18

i like to avoid product rule when i can
and just do simple derivatives where power rule is only thing involved

Answer 19

yeah but you already found the derivative of
\[x\ln(x)\] in the line above so you get it for free! it is not like it changed from one line to the next

Answer 20

lol

Answer 21

it requires more writing though
so it does cost

Answer 22

and no one uses the power rule to find the derivative of
\[\frac{1}{x}\] it is like using a calculator to multiply by 5

Answer 23

i wish zarkon was here so he could agree with me

Answer 24

you could always just delete my post

Answer 25

my way is easiest! i win!

Answer 26

my way is easiest....
you win anyway

Answer 27

good game

Answer 28

you two are too funny

Answer 29

who wins zarkon?

Answer 30

i do of course

Answer 31

no!

Answer 32

(the arrogance!)

Answer 33

a solid tie

Answer 34

no way!

Answer 35

that's what the score card says. I only report what the cards say

Answer 36

rubber match tomorrow

Answer 37

it doesn't matter to you which one: (xln(x))' or (1/x)'
:(
power rule is easiest!

Answer 38

product rule requires more writing

Answer 39

if you know the taylor expansion at x=1 for the ln(x) the solution it pretty short

Answer 40

lol

Answer 41

goodnight zarkon