Question

Can someone help me with this L'Hopital problem: lim as x approaches 1 from the right or (lnx)^(x-1)

Answer 1

I was thinking you could do y=lim.... and then take ln of both sides. But that just makes it more difficult.

Answer 2

take the log first
then do it

Answer 3

yes u have to remove the (x-1) from the exponent first

Answer 4

or else rewrite as
\[\large \ln(x)^{x-1}=e^{(x-1)\ln(\ln(x))}\]

Answer 5

either way the work is identical, finding
\[\lim_{x\to 1^+}(x-1)\ln(\ln(x))\]

Answer 6

\[[ln(x)]^{x-1}=ln^x(x)*ln(x)\]

Answer 7

divide, not multiply

Answer 8

\[[ln(x)]^{x-1}=ln^x(x)~/~ln(x)\]

Answer 9

right?

Answer 10

yeah, but my guess is that it is now going to be even more complicated although i could be way off

Answer 11

was this your suggestion?
\[y=(lnx)^{(x-1)}\]\[e^y=e^{(x-1)~lnx}\]

Answer 12

maybe not though...

Answer 13

no it was just two rewrite \(b^x\) as \(e^{x\ln(b)}\)

Answer 14

usually math teacher says "take the log" and "simplify using properties of the log" then compute the limit and then since you take the log as a first step you "exponentiate" (if that is a word) at the end

Answer 15

so in this problem it would be
\[y=\ln(x)^{x-1}\]
\[\ln(y)=\ln(\ln(x)^{x-1})=(x-1)\ln(x)\]

Answer 16

oops i mean
\[(x-1)\ln(\ln(x))\]

Answer 17

my precalc teacher told us that whenever you see a convoluted setup, the answer is either going to be 0 or 1 :)

Answer 18

lol what about
\[\lim_{x\to \infty}(1+\frac{3}{x})^x\]?

Answer 19

but back to my screed, this is all math teacher nonsense

Answer 20

since by definition
\[b^x:=e^{x\ln(b)}\]

Answer 21

and so by definition
\[\large \ln(x)^{x-1}=e^{(x-1)\ln(\ln(x))}\]

Answer 22

all the work is still computing the limit in the sky however, so it really makes no difference

Answer 23

i wonder about a power series, but that thought really hasnt formed and im sure it would be a nightmare

Answer 24

eeeew

Answer 25

seems nightmareish

Answer 26

\[ln(x)=\sum_1\frac{-(1-x)^n}{n}\]
\[[ln(x)]^{x-1}=\sum_1\left[\frac{-(1-x)^n}{n}\right]^{x-1}\]

Answer 27

yikes!!

Answer 28

oh it aint that bad :) at x=1 that zeros out to 1 i believe

Answer 29

My book kinda of gives me an answer, it say to do:
Let y= lim ln(x)^(x-1),

then somehow its goes to

= lim (x-1)ln(x)

=0

Hence the lim is equal to 1

Answer 30

ok lets go slow

Answer 31

it doesn't go to \((x-1)\ln(\ln(x))\) that is what you get when you take the log

Answer 32

no, its doesnt even say to take ln of both sides

Answer 33

which is unsusal, beacuse the soulutions to the other problems like this one actually say take ln of both sides

Answer 34

ok then maybe i misread the problem
could it be
\[\large \ln(x^{x-1})\]

Answer 35

which is not the same as
\[\large (\ln(x))^{x-1}\] at all

Answer 36

nope, it's