Need help with limite: lim x->inf (ln (e^x -1)) -x)
- Thread starter Jencek007
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What, specifically, did you try? Where are you getting stuck. For instance, you say you "tride [sic] [to apply] L'Hopital's Rule." To use that, the limit in question must be a fraction of an indeterminate form (0/0 or Inf/Inf, etc). So, what steps did you follow trying to convert the given limit to a fraction? If you successfully made a fraction, what was the derivative of the numerator? What was the derivative of the denominator? Please share with us all of your work, even if you know it's wrong. Thank you.
Steven G
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Here is a helpful hint. x = ln(e^x). See where you can go from there.
HallsofIvy
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But that does NOT help with ln(e^x- 1)!
D
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But it does make the original problem manageable!
lim x->inf (ln (e^x -1)) -x)
= lim x->inf [ln (e^x -1)) -ln(e^x)]
= lim x->inf [ln (e^x -1)/(e^x)]
= lim x->inf [ln {1 - e^(-x)}]
and continue.....
I tried it like this
lim x->inf (ln (e^x -1)) -x)
lim x->inf x(ln ((e^x -1)/x) -1)
which means
lim x->inf inf((inf/inf) -1)
l´Hopitales rule for inf/inf
((1/(e^x -1))*e^x)/1 = e^x/(e^x-1) = e^x/(e^x*(1-1/e^x)) = 1/(1-1/e^x)
but when I take this to my limite, it looks like this
lim x->inf x((1/(1-1/e^x)) -1)
and for x->inf it is: inf((1-1) => inf*0
So please can somebody tell me what am I doing wrong?
lim x->inf (ln (e^x -1)) -x)
lim x->inf x(ln ((e^x -1)/x) -1)
which means
lim x->inf inf((inf/inf) -1)
l´Hopitales rule for inf/inf
((1/(e^x -1))*e^x)/1 = e^x/(e^x-1) = e^x/(e^x*(1-1/e^x)) = 1/(1-1/e^x)
but when I take this to my limite, it looks like this
lim x->inf x((1/(1-1/e^x)) -1)
and for x->inf it is: inf((1-1) => inf*0
So please can somebody tell me what am I doing wrong?
I think you went wrong right from the beginning actually. Although I'm finding it a little difficult to follow your work. Here's what I think you did... starting with the given problem, you factored out an x:
\(\displaystyle \displaystyle \lim _{x\to \infty }\left(ln\left(e^x-1\right)-x\right)=\lim _{x\to \infty }\left(x\left(\dfrac{ln\left(e^x-1\right)}{x}-1\right)\right)\)
You then broke it up into three limits, like so:
\(\displaystyle \displaystyle \lim _{x\to \infty }\left(x\left(\frac{ln\left(e^x-1\right)}{x}-1\right)\right)=\lim _{x\to \infty }\left(x\right)\cdot \left(\lim _{x\to \infty }\left(\dfrac{ln\left(e^x-1\right)}{x}\right)-\lim _{x\to \infty }\left(1\right)\right)\)
And then used L'Hopital's rule on the middle limit, which does to go infinity/infinity. But once you did that, you were left with:
\(\displaystyle \lim _{x\to \infty }\left(x\right)\cdot \left(1-1\right)=\infty \cdot 0\)
Which is undefined. Every step along the way was valid, and I didn't spot any math errors, but the steps were probably not the ones you want for this problem. Instead, try going back to the drawing board and follow Subhotosh Khan's hint. Or, as a slightly different method, you might try to assume the limit exists, and call it L.
\(\displaystyle \displaystyle \lim _{x\to \infty }\left(ln\left(e^x-1\right)-x\right)=L\)
Now try "undoing" the natural log by "powering them up" by e. That is to say:
\(\displaystyle \displaystyle \lim _{x\to \infty }\left(e^{ln\left(e^x-1\right)-x}\right)=e^L\)
And then see what comes of that. Remember that add the end, you won't find L by doing so, but rather you'll find e^L, so you'll need to take the natural log to get L, which is the actual value of the limit.
Steven G
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Prof, I am sorry but I think your claim is not correct. Please look at it a bit more closely.
