Question

lim ( (x)/(x-1) - [(1) / (lnx)] )
x->1

Answer 1

i think you have to subtract first

Answer 2

may you please rewrite :D

Answer 3

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

Answer 4

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

Answer 5

\[\lim_{x \rightarrow 1} ( \frac{x}{x-1} - \frac{1}{lnx} )\]

Answer 6

sorry for the confusion

Answer 7

it's a L'H problem

Answer 8

got it
subtract first, then probably l'hopital

Answer 9

so if i plug in 1 for x i get 0/0. so... then what do i do exactly?

Answer 10

i'm suppose to take use L'H, but it's in a weird looking form for that

Answer 11

remember this is the same as
\[(\lim_{x \rightarrow 1}x/x-1)-(\lim_{x \rightarrow 1}1/lnx)\]
\[(\lim_{x \rightarrow 1}1/1)-(\lim_{x \rightarrow 1}0/1/x)\]
then we get 1-0=1
answer is one, sorry my first answer was wrong

Answer 12

oh no

Answer 13

answer is 1/2

Answer 14

\[\lim{x\to 1^+}\frac{1}{\ln(x)}=\infty\] you cannot take the limits separately like that

Answer 15

start with
\[\lim_{x\to 1^+}\frac{x\ln(x)-x+1}{(x-1)\ln(x)}=\lim_{x\to 1^+}\frac{x\ln(x)-x+1}{x\ln(x)-\ln(x)}\] then use l'hopital
take the derivative top and bottom separately

Answer 16

but
\[\lim_{x \rightarrow a}f(x)-g(x)=\lim_{x \rightarrow a}f(x)-\lim_{x \rightarrow a}g(x)\]
right?

Answer 17

derivative of the numerator simplifies to \(\ln(x)\)
@anonymoustwo44 yes, but both of those limits are infinity, so you are in the form \(\infty-\infty\) so you have to do more work

Answer 18

what you wrote is true assuming the limits are finite

Answer 19

@satellite73 ah ok ok

Answer 20

btw there is no + on the 1

Answer 21

ok

Answer 22

i didn't follow what you did :(

Answer 23

i'm looking at x/x-1

Answer 24

in any case if you take the derivative top and bottom separately you get
\[\frac{\ln(x)}{\ln(x)+1-\frac{1}{x}}\] which is still in the form \(\frac{0}{0}\) so you have to do it one more time

Answer 25

take the derivative top and bottom again you get
\[\frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x^2}}\] now replace x by 1 and get
\[\frac{1}{2}\]

Answer 26

it's the minus between the fractions i dont know how to deal with...

Answer 27

back up to the first step

Answer 28

\[\lim_{x \rightarrow 1} ( \frac{x}{x-1} - \frac{1}{\ln(x)} )=\lim_{x\to 1}\frac{x\ln(x)-x+1}{(x-1)\ln(x)}=\lim_{x\to 1}\frac{x\ln(x)-x+1}{x\ln(x)-\ln(x)}\] this step is pure algebra, just subtraction

Answer 29

next step is take the derivative top and bottom
you get
\[\lim_{x\to 1}\frac{\ln(x)}{\ln(x)+1-\frac{1}{x}}\]

Answer 30

try replacing x by 1 and you see that you get \(\frac{0}{0}\)

Answer 31

what's happening in that first step

Answer 32

this one
\[\lim_{x \rightarrow 1} ( \frac{x}{x-1} - \frac{1}{\ln(x)} )=\lim_{x\to 1}\frac{x\ln(x)-x+1}{(x-1)\ln(x)}=\lim_{x\to 1}\frac{x\ln(x)-x+1}{x\ln(x)-\ln(x)}\]??

Answer 33

yes.. after the first =

Answer 34

i know it's algebra.. that's my problem :(

Answer 35

that is algebra
\[\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bc}\]

Answer 36

i think you mean over bd

Answer 37

yes right

Answer 38

multiply the two denominators together to get
\[(x-1)\ln(x)=x\ln(x)-\ln(x)\] for our denominator

Answer 39

numerator is
\[x\ln(x)-(x-1)=x\ln(x)-x+1\]

Answer 40

ok with ya

Answer 41

so that gives
\[\frac{x\ln(x)-x+1}{x\ln(x)-\ln(x)}\] right ?

Answer 42

yes

Answer 43

now we want to replace x by 1 but we cannot because we will get \(\frac{0}{0}\)

Answer 44

yes

Answer 45

so we use l'hopital's rule and take the derivative of the top and bottom. that will give
\[\lim_{x\to 1}\frac{\ln(x)}{\ln(x)+1-\frac{1}{x}}\]

Answer 46

we again try to replace x by 1 and again we get \(\frac{0}{0}\)

Answer 47

we did product rule in numerator, right?

Answer 48

well both i guess

Answer 49

so we have to use l'hopital one more time
this time we get
\[\frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x^2}}\]

Answer 50

yes product rule in numerator and denominator to find the derivative of \(x\ln(x)\)

Answer 51

kk

Answer 52

then that's 1/2... got it. shesh!

Answer 53

btw, how do you type out the numbers so fast? it takes me forever to use that thing

Answer 54

i am not using the equation editor

Answer 55

what's your secret

Answer 56

i am typing in latex

Answer 57

so for example if i want
\[\lim_{x\to 1}\frac{\ln(x)}{\ln(x)+1-\frac{1}{x}}\] i type in
\lim_{x\to 1}\frac{\ln(x)}{\ln(x)+1-\frac{1}{x}}

Answer 58

by hand? good grief charlie brown

Answer 59

i started answering questions here to practice my latex almost a year ago, and know i am getting better at it, but i got hooked on answering qustions
you enclose the code by first typing "\["

Answer 60

and end it with "\]"
try it if you like
if you want to see the code for anything written, right click on it and select "view latex"

Answer 61

why would you want to get good at latex, is that a valued skill anywhere?

Answer 62

you can try for example \int_0^{\infty}e^{-x}dx and see what you get

Answer 63

no it is only good for type setting

Answer 64

it is the way one writes math, nothing further

Answer 65

/[\int_0^{\infty}e^{-x}dx]

Answer 66

oops

Answer 67

oh well i'll work on it.. ty for the help!