# [SOLVED] stuck on a limit and out of ideas

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• Mar 5th 2009, 12:25 PM
sinewave85
[SOLVED] stuck on a limit and out of ideas
I can't get this limit into any form that does not involve an undefined quantity -- specifically ln(0) or 0^(0).

$\displaystyle \lim_{x\rightarrow0}(e^{x} - 1 - x)^{x}$

This does not work, obviously:

$\displaystyle \ln L = \lim_{x\rightarrow0}\frac{\ln(e^{x} - 1 - x)}{1/x}$

I have fiddled around unproductively with equivalent equations trying to get it into a form suitable to l'Hopital's equation, but I am not going type all of that out since it came to naught. Any help or hints would be appreciated.
• Mar 5th 2009, 12:45 PM
Prove It
Quote:

Originally Posted by sinewave85
I can't get this limit into any form that does not involve an undefined quantity -- specifically ln(0) or 0^(0).

$\displaystyle \lim_{x\rightarrow0}(e^{x} - 1 - x)^{x}$

This does not work, obviously:

$\displaystyle \ln L = \lim_{x\rightarrow0}\frac{\ln(e^{x} - 1 - x)}{1/x}$

I have fiddled around unproductively with equivalent equations trying to get it into a form suitable to l'Hopital's equation, but I am not going type all of that out since it came to naught. Any help or hints would be appreciated.

Hint: $\displaystyle \lim_{x \to 0} {x^x} = 1$
• Mar 5th 2009, 01:27 PM
sinewave85
Quote:

Originally Posted by Prove It
Hint: $\displaystyle \lim_{x \to 0} {x^x} = 1$

Ok, I am in dumb mode I guess, but I still can't get it. Obvioulsy the limit approximates $\displaystyle \lim_{x \to 0} {x^x} = 1$, but I am still not able to get to a form that works. Any help is appreciated!
• Mar 5th 2009, 01:34 PM
Moo
Hello,

Maybe you can note that when x is near 0, $\displaystyle e^x \sim x$ (is equivalent to x)

so we can have $\displaystyle (e^x-1-x)^x \quad \sim_{x \to 0} \quad (x-1-x)^x=(-1)^x$

not too sure about that :s
• Mar 5th 2009, 01:49 PM
sinewave85
Quote:

Originally Posted by Moo
Hello,

Maybe you can note that when x is near 0, $\displaystyle e^x \sim x$ (is equivalent to x)

so we can have $\displaystyle (e^x-1-x)^x \quad \sim_{x \to 0} \quad (x-1-x)^x=(-1)^x$

not too sure about that :s

But wouldn't $\displaystyle e^x \to 1$ and not 0 as as $\displaystyle x \to 0$?
• Mar 5th 2009, 01:54 PM
sinewave85
The other thing is, running the expression through a calculator, simply checking very small numbers, it appears that the limit is not, in fact, 1 but 0, so we may be on the wrong track entirely.
• Mar 5th 2009, 03:05 PM
Prove It
Quote:

Originally Posted by sinewave85
The other thing is, running the expression through a calculator, simply checking very small numbers, it appears that the limit is not, in fact, 1 but 0, so we may be on the wrong track entirely.

Notice what happens when $\displaystyle x \to 0$.

$\displaystyle e^x \to 1, x \to 0$ so $\displaystyle e^x - 1 - x \to 0$.

So ultimately it tends to $\displaystyle 0^0$, which tends to 1.

Correct me if my logic is wrong.
• Mar 5th 2009, 05:54 PM
sinewave85
Quote:

Originally Posted by Prove It
Notice what happens when $\displaystyle x \to 0$.

$\displaystyle e^x \to 1, x \to 0$ so $\displaystyle e^x - 1 - x \to 0$.

So ultimately it tends to $\displaystyle 0^0$, which tends to 1.

Correct me if my logic is wrong.

It seems like that should be right, and I don't question your logic; the only thing is that, for reasons far beyond me, this expression does not seem to tack with the general principal you reminded me of. For example, let x = 0.000000001. Running it through a calculator:

$\displaystyle (e^{x} - 1 - x)^{x} = 0$

but

$\displaystyle x^{x} = 0.9999998158 \approx 1$

I just can not get an answer to this limit logically that matches what I get by simply solving the expression for values very close to
x = 0. It is from a lesson on l'Hopital's rule, but not every problem in the set requires its use. No mater which approach I have taken, I either end up with an undefined expression not suitable to l'Hopital's rule or a false answer. Once again, any help would be greatly appreciated.
• Mar 5th 2009, 10:15 PM
Moo
I'm sorry, I guess I was very tired yesterday, to write such a thing.

But I don't agree with Prove_it's method, since it's not exactly x^x here... but it looks like it gives the correct result

I applied twice (or three times) l'Hopital's rule :

Let's start like you did
$\displaystyle \ln(L)=\lim_{x \to 0} ~ \frac{\ln(e^x-x-1)}{1/x}$
this is in the form $\displaystyle \frac{\infty}{\infty}$, because :
o $\displaystyle e^x-x-1 \to 0 \Rightarrow \ln(\dots) \to -\infty$
o $\displaystyle 1/x \to \infty$

So we can apply l'Hopital's rule :
$\displaystyle \ln(L)=\lim_{x \to 0} ~ \frac{\frac{e^x-1}{e^x-x-1}}{-1/x^2}=\lim_{x \to 0} ~ -x^2 \cdot \frac{e^x-1}{e^x-x-1}$

now note that $\displaystyle \frac{e^x-1}{e^x-x-1}=1+\frac{x}{e^x-x-1}$
hence
$\displaystyle \ln(L)=\lim_{x \to 0} ~ -x^2-\frac{x^3}{e^x-x-1}$

now apply l'Hopital's rule for the right term, since it's $\displaystyle \frac 00$ :
$\displaystyle \ln(L)=\lim_{x \to 0} ~ -x^2-\frac{3x^2}{e^x-1}$

apply it again :
$\displaystyle \ln(L)=\lim_{x \to 0} ~ -x^2-\frac{6x}{e^x}$

and this limit is obviously 0.

--> L=1
• Mar 6th 2009, 06:14 AM
Soroban
Hello, sinewave85!

Quote:

$\displaystyle \lim_{x\to 0}(e^x - 1 - x)^x$

This does not work: .$\displaystyle \ln L \:= \:\lim_{x\to0}\frac{\ln(e^x - 1 - x)}{x^{-1}}$

Did you apply L'Hopital?

We get: .$\displaystyle \frac{\dfrac{e^x-1}{e^x - x - 1}} {-x^{-2}} \;=\;\frac{-x^2(e^x-1)}{e^x-x-1} \quad\to\quad\frac{-0(1-1)}{1-0-1} \;=\;\frac{0}{0}$

Apply L'Hopital again: .$\displaystyle \frac{-x^2e^x - 2xe^x + 2x}{e^x-1} \quad\to\quad \frac{-0-0+0}{1-1} \;=\;\frac{0}{0}$

Apply L'Hopital again: .$\displaystyle \frac{-x^2e^x - 4xe^x - 2e^x + 2}{e^x} \quad\to\quad \frac{-0+0-2+2}{1} \:=\:0$

. . We have: .$\displaystyle \ln L \:=\:0$

. . Therefore: .$\displaystyle L \;=\;e^0 \;=\;1$

• Mar 6th 2009, 07:12 AM
sinewave85
Thanks Moo and Soroban! I got so tangled up in this one that I could not see a way out. I still wonder why the calculator approzimation is not "true," but the solution is obviously 1, so I guess I can live with the uncertainty.
• Mar 6th 2009, 08:23 AM
Opalg
Quote:

Originally Posted by sinewave85
I still wonder why the calculator approzimation is not "true," but the solution is obviously 1, so I guess I can live with the uncertainty.

Calculators can only work to a certain number of decimal places. When you ask the calculator to evaluate $\displaystyle (e^x-1-x)^x$ with $\displaystyle x=10^{-9}$ say, it starts by evaluating $\displaystyle e^x-1-x$ for that value of x. But $\displaystyle e^x = 1+x+\tfrac12x^2+$ higher powers of x, so $\displaystyle e^x-1-x$ is approximately $\displaystyle x^2/2$, which in this case will be less than $\displaystyle 10^{-18}$. That is probably too small a number for the calculator to store, so the poor machine suffers from "arithmetic underflow" and replaces this quantity by 0. Then when 0 is raised to any power, it is still zero, of course. So the calculator gives the result of the calculation as 0.

If you start with somewhat larger values of x, say $\displaystyle x=10^{-1},\ 10^{-2},\ 10^{-3},\ \ldots,$ you should find that the values that the calculator gives for $\displaystyle (e^x-1-x)^x$ get closer and closer to 1, until at a certain stage arithmetic underflow sets in and the values suddenly become 0.
• Mar 6th 2009, 09:41 AM
sinewave85
Quote:

Originally Posted by Opalg
Calculators can only work to a certain number of decimal places. When you ask the calculator to evaluate $\displaystyle (e^x-1-x)^x$ with $\displaystyle x=10^{-9}$ say, it starts by evaluating $\displaystyle e^x-1-x$ for that value of x. But $\displaystyle e^x = 1+x+\tfrac12x^2+$ higher powers of x, so $\displaystyle e^x-1-x$ is approximately $\displaystyle x^2/2$, which in this case will be less than $\displaystyle 10^{-18}$. That is probably too small a number for the calculator to store, so the poor machine suffers from "arithmetic underflow" and replaces this quantity by 0. Then when 0 is raised to any power, it is still zero, of course. So the calculator gives the result of the calculation as 0.

If you start with somewhat larger values of x, say $\displaystyle x=10^{-1},\ 10^{-2},\ 10^{-3},\ \ldots,$ you should find that the values that the calculator gives for $\displaystyle (e^x-1-x)^x$ get closer and closer to 1, until at a certain stage arithmetic underflow sets in and the values suddenly become 0.

Thank you very much for the thorough explanation. Not knowing why the discrepancy occured was still bothering me. I will be more carefull about watching for overflow from now on.
• Mar 6th 2009, 11:21 PM
CaptainBlack
Quote:

Originally Posted by Prove It
Notice what happens when $\displaystyle x \to 0$.

$\displaystyle e^x \to 1, x \to 0$ so $\displaystyle e^x - 1 - x \to 0$.

So ultimately it tends to $\displaystyle 0^0$, which tends to 1.

Correct me if my logic is wrong.

The answer is right but the logic is wrong $\displaystyle 0^0$ is undefined as what you get for

$\displaystyle L=\lim_{x \to 0} ((f(x)^{g(x)})$

with $\displaystyle f$ and $\displaystyle g$ continuous at $\displaystyle 0$ and $\displaystyle f(0)=g(0)=0$ depends how $\displaystyle f$ and $\displaystyle g$ behave near $\displaystyle 0$.

CB
• Mar 7th 2009, 09:34 AM
sinewave85
Quote:

Originally Posted by CaptainBlack
The answer is right but the logic is wrong $\displaystyle 0^0$ is undefined as what you get for

$\displaystyle L=\lim_{x \to 0} ((f(x)^{g(x)})$

with $\displaystyle f$ and $\displaystyle g$ continuous at $\displaystyle 0$ and $\displaystyle f(0)=g(0)=0$ depends how $\displaystyle f$ and $\displaystyle g$ behave near $\displaystyle 0$.

CB

That is good to know, too. I might have been tempted to assume that any function that approximates $\displaystyle \lim_{x\to0} = x^{x}$ behaves accordingly.
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