Question

limit as x goes to 1+
(x^2-1)^(x-1)
using l'hopitals rule

Answer 1

I no of no other way to do this than to exponentiate this first

Answer 2

\[\lim_{x\to a}f(x)=e^{\lim_{x\to a}\ln f(x)}=\exp\left(\lim_{x\to a}\ln f(x)\right)\]do you know this trick?

Answer 3

so (x-1)ln(x^2-1)

Answer 4

\[(x-1)\ln(x^2-1)\]

Answer 5

yes, and now I would do the 1/t sub
I bet there is a better way though... that anonymous is about to show us

Answer 6

ok lol ill wait for his response

Answer 7

oh no, my sub would be wrong

Answer 8

maybe let x-1=1/t and let t go to infinity

Answer 9

\[(x-1)[\ln(x-1)+\ln(x+1)]={[\ln(x-1)+\ln(x+1)]\over(x-1)^{-1}}\]I am very stuck here

Answer 10

Zarkon to the rescue!

Answer 11

do you know that \[\lim_{x\to 0^+}x^x=1\]?

Answer 12

ooooooh

Answer 13

\[\lim_{x\to1^+}(x-1)^{x-1}(x+1)^{x-1}=\lim_{x\to1^+}(x+1)^{x-1}\]that is a step in the right direction I suppose

Answer 14

yep...and the answer follows

Answer 15

\[2^0=1\]

Answer 16

oh yeah, lol it's defined now

Answer 17

\[\lim_{x \rightarrow 1^+}(x^2-1)^{(x-1)}=\lim_{x \rightarrow 1^+}(x^2-1)^x/(x^2-1)\]
lim (x^2-1)(x^2-1)(x^2-1)......(x^2-1)/(x^2-1)
x-> 1^+ ^ x times
lim (x^2-1)(x^2-1).....(x^2-1)
x->1^+ ^ x-1 times
=(1-1)(1-1).....(1-1)=0

Answer 18

...but the limit is 1

Answer 19

isn't 0^0 undefined?

Answer 20

sorry to ask

Answer 21

\[0^0\] is undefined...but \[\lim_{x\to 0^+}x^x\] is not

Answer 22

not undefined that is ;)

Answer 23

remember in the limit x is never 0

Answer 24

\[\lim_{x\to0^+}x^x=\lim_{x\to0^+}\exp(x\ln x)=\exp(\lim_{t\to\infty}{-\ln t\over t})=\exp(\lim_{t\to\infty}-\frac1t)\]\[=\exp(0)=e^0=1\]as @Zarkon said

Answer 25

oh ok I forgot product rule

Answer 26

thanks everyone wish you guys could all be my study buddies! :D

Answer 27

we are ;)
welcome!

Answer 28

i mean like in person study buddies lol!