Question

Use l'Hopital's Rule to evaluate lim(x−>1)(x2−2x+1)^(x−1) and identify the indeterminate form

Answer 1

my big confusion is why ln(0)=positive infinity

Answer 2

ln(x->0)

Answer 3

ln(x->0+)

Answer 4

That's not the kind of thing that you solve with l'Hospital's rule. That rule can only be applied to fractions.

Answer 5

But you can solve it with L'hopitals if you take the ln of the inside part

Answer 6

well the ln of the thing you are finding the limit of

Answer 7

Then you get something that you can rewrite, and then do l'hopitals with

Answer 8

That's how 0^0 is solved, yeah. I don't think it's called l'Hospital's rule though. To my knowledge at least.

Answer 9

You do apply l'hopitals in the problem. That particular step is not l'hopitals though

Answer 10

ANYWAYS, why is the limit of ln(x) as x->0+ positive infinity

Answer 11

it's not

Answer 12

According to these solutions I found it is

Answer 13

\[\lim_{x\to 0^{+}}\ln(x)=-\infty\]

Answer 14

Yes

Answer 15

Why

Answer 16

don't believe everything you read.

Answer 17

Wait no

Answer 18

graph ln(x)

Answer 19

I know what ln(x) looks like

Answer 20

If you think about the definition of a cusp, one limit goes to infinity, and one goes to negative infinity, even though it does look like both go to one infinity

Answer 21

what are you talking about

Answer 22

http://www.sosmath.com/calculus/diff/der09/der09.html

Answer 23

Scroll down to the part where it defines a cusp

Answer 24

there are no cusps for ln(x)

Answer 25

I know that, but I think that it is the exact same case. Since one of the cusp's limits goes to positive infinity, I think that is the same thing for ln(x), but idk why it is positive

Answer 26

Here's the solution I found http://www.slader.com/textbook/9780132014083-calculus-graphical-numerical-algebraic-3rd-edition/450/exercises/23/

Answer 27

They say that it is positive infinity over positive infinity

Answer 28

and what does that have to do with just ln(x)

Answer 29

those are more complicated functions

Answer 30

ln(x) is like half of a cusp

Answer 31

it is not

Answer 32

And I guess it is the half that goes to positive infinity

Answer 33

In terms of limits it is

Answer 34

cusps have a definite stopping point



the ln(x) does not