Thread: Question on Limits(with intergration)

1. Question on Limits(with intergration)

my question is why does the book say:

when u plugs in 1- in that.. u get 5(0)^4/5.. which is zero.. even though its 1- its still 0..

but here if u plug in o- to the e^x... u would get 1-1 = ln0 which does not exsist.. the book says the answer is infinite

i guess more speciafically my question is why does the first problem above come out to 1-1 = 0 even though its 1-... and on the 2nd equation u gotaa use the 0- as a actual 0- to get infinite..

any help.. sorry i know this question may not make sense.

2. I'm not quite understanding your question but ...

$\lim_{t \to 1^{-}} \left[5(x-1)^{\frac{4}{5}}\right]_{0}^{t}
\: = \: \lim_{t \to 1^{-}} \left(5(t - 1)^{\frac{4}{5}} - 5(0-1)^{\frac{4}{5}}\right) = 5(0-1)^{\frac{4}{5}} - 5(-1)^{\frac{4}{5}} = 5 - 5 = 0$

$\lim_{t \to 0^{-}} \bigg[ \ln \left| e^{x} - 1 \right| \bigg]_{0}^{t} = \lim_{t \to 0^{-}} \bigg( \ln \left| e^{t} - 1\right| - \ln \left| e^{0} - 1\right| \bigg) = \left(\ln 0 - \ln 0\right) = -\infty - (- \infty)$

The latter is an indeterminate form but it does converge.

As for the $1^{-}$, you don't really need it in this case but there are cases in which the limit as you approach a number from the left side is different from approaching it the right side.

3. Originally Posted by o_O
I'm not quite understanding your question but ...

$\lim_{t \to 1^{-}} \left[5(x-1)^{\frac{4}{5}}\right]_{0}^{t}
\: = \: \lim_{t \to 1^{-}} \left(5(t - 1)^{\frac{4}{5}} - 5(0-1)^{\frac{4}{5}}\right) = 5(0-1)^{\frac{4}{5}} - 5(-1)^{\frac{4}{5}} = 5 - 5 = 0$

$\lim_{t \to 0^{-}} \bigg[ \ln \left| e^{x} - 1 \right| \bigg]_{0}^{t} = \lim_{t \to 0^{-}} \bigg( \ln \left| e^{t} - 1\right| - \ln \left| e^{0} - 1\right| \bigg) = \left(\ln 0 - \ln 0\right) = -\infty - (- \infty)$

The latter is an indeterminate form but it does converge.

As for the $1^{-}$, you don't really need it in this case but there are cases in which the limit as you approach a number from the left side is different from approaching it the right side.
so you mean to say ln(0) = -infinity??

i thought ln(0)= does not exsist...

and the book says the answer for that 2nd one is infinity..

and u got -infinity+infinity... how did they get possitive infintiy from that?

4. Originally Posted by Legendsn3verdie
so you mean to say ln(0) = -infinity??

i thought ln(0)= does not exsist...

and the book says the answer for that 2nd one is infinity..

and u got -infinity+infinity... how did they get possitive infintiy from that?
It's an indeterminate form that you cannot use L'hopital on..you have to change it into a form like $\frac{\infty}{\infty}$ or $\frac{0}{0}$ I think in order to use L'Hopital to get the correct result.

I think.

5. Originally Posted by elizsimca
It's an indeterminate form that you cannot use L'hopital on..you have to change it into a form like $\frac{\infty}{\infty}$ or $\frac{0}{0}$ I think in order to use L'Hopital to get the correct result.

I think.

i see.. how do u know its inderteminate????

6. Well ln(0) doesn't exist but as you approach it, it will tend to negative infinity which is what we mean by the limit being negative infinity.

As for that last one, I'm PRETTY sure it converges but ... let's see:

Edit: Ok I thought the lower limit was 0 instead of -1. My bad. So never mind about l'hopital's rule:

$\lim_{t \to 0^{-}} \bigg( \ln \left| e^{t} - 1\right| - \ln \left| e^{-1} - 1\right| \bigg) =\lim_{t \to 0^{-}} \ln \left(\frac{|e^{t} - 1| }{|e^{-1} - 1|}\right)$

For this one, if we directly plug in 0, we would get negative infinity. But as I graph it, it seems to converge to about -27. Maybe it's just my eyes and it does in fact tend off to negative infinity.