Question

Will someone please help me use L'hopital to evaluate the limit as x-->0+ x^x-1/lnx+x-1

Answer 1

\[\lim_{x \rightarrow o+} x^x-1/lnx+x-1\]

Answer 2

I can't see how this would be or could be re-written in the indeterminate form...

Answer 3

hmm i don't see it either ... i know the limit is 0 though

Answer 4

how do you know it is zero?

Answer 5

plug in consecutive small values for x ... it seems to go toward 0
also http://www.wolframalpha.com/input/?i=lim+%28x^x-1%29%2F%28log%28x%29%2Bx-1%29+as+x-%3E0%2B
:)

Answer 6

i am not sure what the expression is
is it
\[\frac{x^x-1}{\ln(x)+x-1}\]?

Answer 7

if so you cannot use l'hopital because it is not in the form \(\frac{0}{0}\)
the form would be \(\frac{-1}{-\infty}\)

Answer 8

do I rewrite it then? or solve another way?

Answer 9

it is zero

Answer 10

no work, just your eyeballs
numerator approaches -1, denominator goes to \(-\infty\)

Answer 11

so the limit as x approaches 0+ of x^x-1 is -1 because as x^x is basically zero. And therefore x^x -1 is -1?

Answer 12

yes

Answer 13

ooooh no no no sorry that is wrong!!

Answer 14

my mistake

Answer 15

thank you. Our class moves so fast I don't always absorb stuff well...that makes sense....and -1 over infinity is essentially zero

Answer 16

\[\lim_{x\to 0}x^x=1\]

Answer 17

i made a mistake sorry

Answer 18

the limit in the numerator is \(0\) not \(-1\)

Answer 19

so I have to re write the equation, so I can use L'hopitals rule?

Answer 20

but that doesn't make any difference to the solution, because as \(x\to 0\) we have
\[\ln(x)\to -\infty\]

Answer 21

no you cannot use l'hopital, because the denominator does not go to zero

Answer 22

if it was \(\frac{0}{0}\) you could use l'hopital, but it is not

Answer 23

the numerator goes to zero
the denominator goes to \(-\infty\)

Answer 24

so how do I evaluate this?

Answer 25

\(\frac{0}{\infty}=0\) it is not indeterminate

Answer 26

you have a fraction
the numerator is getting closer and closer to zero, but the denominator is going to \(-\infty\)

Answer 27

ok...again..you make sense...so it is just zero then for the answer?

Answer 28

the ratio is getting smaller very fast

Answer 29

yes, zero is the answer
if the denominator was also going to zero, you would have more work to do, but since the denominator is going to minus infinity, you have nothing to do

Answer 30

oh ok...I think I almost get it...the zero isn't really a zero...it is just essentially a zero because it is a number getting more and more tiny...so a number over infinity is essentially zero, which is the answer????

Answer 31

i think i see the confusion
\[\frac{0}{\infty}=0\] but also \(\frac{a}{\infty}=0\)

Answer 32

they are both zero in the limit
if the numerator goes to a finite number, and the denominator goes to infinity, then the limit is zero

Answer 33

for example
\[\lim_{x\to 0^+}\frac{100}{\ln(x)}=0\]

Answer 34

got it...thank you :)