Applications of L'Hopital

[sisxixon]

I'm stuck on finding the limit as x approaches positive infinity x[ (1+ 1/x)^x - e ]


what i've got so far is to let u=1/x, and finding the limit as u --> 0 instead, applied L'Hopital to get

lim u--> 0 Du (1+u)^(1/u) = lim u-->0 [(1/(u+u^2)-ln(1+u)/u^2)](1+u)^(1/u), = infinity.

problem is, the limit is supposed to be -e/2

i know i've done something wrong, or should go a few steps further, but i'm at a loss as to what i should do. suggestions, please?

 

[stapel]

Are you sure the limit value is supposed to be "-e/2"? I mean, isn't the limit of the first bit equal to e itself, so the difference would be zero?

I may be remembering this wrong, but I'd thought the following was true...?


. . . . .\(\displaystyle \large{\begin{array}{c}lim\\x\rightarrow \infty\end{array}\;\left(1\, +\, \frac{1}{x}\right)^x\,=\,e}\)


I could be wrong, of course....

Eliz.

 

[Gene]

One definition of e is
lim(1+1/x)^x as x -> oo
so your limit is zero, not -e/2

(As always Eliz is faster (and corrector)

 

[sisxixon]

the book's got what you've posted, and if i apply that to the original equation, i get
lim x--> infinity = 1/0 (e-e) = 0/0, which means i've got to apply L'Hopital.

heh, i'm still stuck there. do you, by any chance, see where i've gone wrong?

[edit]
the book answer's -e/2, by the way, and if you graph the equation it is -e/2 as you go to infinity.. :shock:

 

[stapel]

Where is the "1/0" coming from?

You've got the "1 + (1/x)", raised to the power x, which goes to "e", and then you're subtracting "e" from this. How are you getting this other "1/0" stuff?

Eliz.

 

[sisxixon]

i've also got the x [ (1+ 1/x)^x - e ] part, and oo=1/0

my problem is it'll still be indeterminate if i just plug infinity in, so i let u=1/x and then used L'Hopital (apologies for being redundant, i know i'm not the most articulate of people and am trying to make this easier on both of you-- thanks for the help so far, by the way)

and after that, i'm stuck.

 

[stapel]

Oh; so the limit is actually "x {[1 + (1/x)]^x - e}"? Then I still don't see where you're getting "1/0" from. The curly-brace part is zero, and the x goes to infinity, and \(\displaystyle 0\,\cdot\,\infty\) does not equal 1/0 or infinity.

Please show all your steps (like we did -- I'm afraid I can't follow the stream-of-conscious computing shown earlier) so we can study what you're doing and how you're getting your answer. Thank you.

Eliz.

 

[sisxixon]

here are my steps:

lim x-->oo x {[1 + (1/x)]^x - e}
1. plug oo in, get oo{e-e}=oo * 0

since this is an indeterminate form, use L'Hopital to find the limit.

here, i let u=1/x so that L'Hopital can be applied, and took the limit u-->0 instead because 1/oo=0

lim u-->0 1/u {[1 + u]^(1/u) - e}

by L'Hopital, take the derivative of the top, and the derivative of the bottom, resulting in

lim u-->0 [(1/(u+u^2)-ln(1+u)/u^2)](1+u)^(1/u)

plug u=0 in and i'll get (oo-oo)e

now, if i apply L'Hopital again, i still won't be able to find the limit as u-->0, although my calculator and textbook both say the limit is -e/2.

 

[soroban]

Hello, sisxixon!

It takes persistence . . . and some stamina.

\(\displaystyle \L\lim_{u\to0}\left[\frac{1}{u(1\,+\,u)} \,-\,\frac{\ln(1\,+\,u)}{u^2}\right]\cdot(1\,+\,u)^{\frac{1}{u}}\;\;\) . . . You did great!

We have: \(\displaystyle \L\:\left[\frac{u\,-\,(1+u)\cdot\ln(1+u)}{u^2(1+u)}\right]\cdot(1 + u)^{1/u}\)


The fraction goes to \(\displaystyle \frac{0}{0}\), so we apply L'Hopital again . . .

\(\displaystyle \L\;\;\left[\frac{1\,-\,(1+u)\cdot\frac{1}{1+u}\,-\,1\cdot\ln(1+u)}{3u^2\,+\,2u}\right]\cdot(1\,+\,u)^{\frac{1}{u}}\\)

\(\displaystyle \L\;\;=\;\left[\frac{-\ln(1+u)}{u(3u\,+\,2)}\right]\cdot(1\,+\,u)^{\frac{1}{u}}\)


Again, the fraction goes to \(\displaystyle \frac{0}{0}\), so we apply L'Hopital again . . .

\(\displaystyle \L\;\;\left[\frac{-\frac{1}{1+u}}{6u\,+\,2}\right]\cdot(1\,+u)^{\frac{1}{u}}\;=\;\frac{-(1\,+\,u)^{\frac{1}{u}}}{2(3u\,+\,1)(1\,+\,u)}\)


Now, apply the limit: \(\displaystyle \L\;\lim_{u\to0}\left[\frac{-(1\,+\,u)^{\frac{1}{u}}}{2(3u\,+\,1)(1\,+\,u)}\right]\)

\(\displaystyle \L\;\;\;=\;\frac{e}{2\cdot1\cdot1}\;= \;-\frac{e}{2}\)