$\lim_{x \to \infty} x-x^2\ln(1+1/x)$ I tried to write it as a fraction and use L'Hospital Rule but it will still be a indetermination. The answer seems to be $1/2$. Can somebody help me, please?
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What is a nedetermination? – gd1035 Oct 25 '18 at 19:39
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Indeterminate probably. – KM101 Oct 25 '18 at 19:40
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Do you know about Taylor series? Can you represnt $\ln (1+ x)$ as a series? Now replace $x$ with $\frac1x$ in that series. – Doug M Oct 25 '18 at 19:41
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Searching in Approach0 for $\lim_{x \to \infty} x-x^2\ln(1+1/x)$ and for $\lim_{x \to \infty} (x-x^2\ln(1+1/x))$ returns several posts about the same limit. (In general, it is a reasonable advice to search first whether your question has not already been answered - although I am fully aware that searching on this sight is not always easy.) – Martin Sleziak Oct 27 '18 at 08:48
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@MartinSleziak That's the first information which should be given to the new users in order to avoid useless questions. But we shouldn't give the duplicates containing full answer otherwise the problem will never be solved. I appreciate a lot your work and effort here. – user Oct 27 '18 at 08:57
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To list some of those questions specifically: 1702282, 1150431, 2003300, 1998703, 1538524. – Martin Sleziak Oct 27 '18 at 09:01
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@MartinSleziak I think that your guidelines How to search on this site? should be implemented in such way that all new users are aware about that and forced to search for duplicates if they are aimed simply to find a solution (and here we could discuss if in that way we are not helping to solve homework too, but I don't want to discuss that here). – user Oct 27 '18 at 09:02
5 Answers
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HINT
To use l'Hopital let consider $$x-x^2\ln\left(1+\frac1x\right)=\frac{\frac1x-\ln\left(1+\frac1x\right)}{\frac1{x^2}}$$
then take $y=\frac1x \to 0^+$ to obtain
$$\lim_{x\to \infty} x-x^2\ln\left(1+\frac1x\right)=\lim_{y\to 0^+} \frac{y-\ln\left(1+y\right)}{y^2}$$
user
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@egreg I suppose that the asker is not aware about that! To avoid l'Hopital we can also use L3 derivation here. – user Oct 26 '18 at 23:06
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You still need to use l'Hopital twice here, which is fine, but it might not be obvious to everyone. – The Count Oct 27 '18 at 04:34
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@TheCount Intentionally I didn't give a full solution for many reasons, notably: it seems a homework, it is a simple problem, the asker knows already the result and is asking for an help to apply l'Hopital, the asker didn't show his work on that, the asker can find the solution in other duplicates, etc. In these cases give a full solution is not good for the asker nor for the site. Moreover since the asker accepted the answer I suppose he/she as successfully solved it by him/herself and that a good thing. If the asker or anyone is interested in more detail I'm at disposal for clarifications. – user Oct 27 '18 at 06:28
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@user why is it wrong to argue like \begin{align} \lim_{x \to \infty} \left[x - x ^ 2 \cdot \log\left(1 + \frac{1}{x}\right)\right] & = \lim_{x \to \infty} x - \lim_{x \to \infty} x ^ 2 \cdot \lim_{x \to \infty} \log\left(1 + \frac{1}{x}\right) \ & = \infty - 0 = \infty ? \ \end{align} – johnny09 Dec 06 '19 at 00:06
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Using the integral definition of the natural logarithm and a simple property of the greatest integer function, we have
$$x-x^2\ln(1+1/x)=x-x^2(\ln(x+1)-\ln(x))=x^2\int_x^{x+1}{dt\over\lfloor t\rfloor}-x^2\int_x^{x+1}{dt\over t}=x^2\int_x^{x+1}{t-\lfloor t\rfloor\over t\lfloor t\rfloor}dt$$
Now
$$\int_x^{x+1}{t-\lfloor t\rfloor\over t\lfloor t\rfloor}dt\le{1\over x(x-1)}\int_x^{x+1}(t-\lfloor t\rfloor)dt$$
and
$$\int_x^{x+1}{t-\lfloor t\rfloor\over t\lfloor t\rfloor}dt\ge{1\over (x+1)x}\int_x^{x+1}(t-\lfloor t\rfloor)dt$$
Finally, $t-\lfloor t\rfloor$ is periodic with period $1$, so
$$\int_x^{x+1}(t-\lfloor t\rfloor)dt=\int_0^1(t-\lfloor t\rfloor)dt=\int_0^1t\,dt={1\over2}$$
Thus
$${x\over2(x+1)}\le x-x^2\ln(1+1/x)\le{x\over2(x-1)}$$
and so by the Squeeze Theorem we have
$$\lim_{x\to\infty}(x-x^2\ln(1+1/x))={1\over2}$$
Barry Cipra
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That's really an original approah! Never saw that, I need to look it better. – user Oct 27 '18 at 06:31
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hint
$$\ln(1+X)=X-\frac{X^2}{2}(1+\epsilon(X))$$
with $$\lim_{X\to 0}\epsilon(X)=0$$
think $X\to \frac 1x$.
hamam_Abdallah
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Maybe the asker is not aware aboyt Taylor, indeed he is asking for a way to apply l'Hopital but it is a good hint for the future! – user Oct 27 '18 at 06:29
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Denote: $1+\frac1x=e^y$. Then: $$\lim_{x \to \infty} x-x^2\ln(1+1/x)=\lim_{y \to 0^+} \frac{1}{e^y-1}-\frac{y}{(e^y-1)^2}=\lim_{y \to 0^+} \frac{e^y-1-y}{(e^y-1)^2}\stackrel{LR}{=}\\ \lim_{y \to 0^+} \frac{e^y-1}{2(e^y-1)e^y}=\frac12.$$
farruhota
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Thank you. It is a slight variation of your solution: if you let $1+y=e^t$, you get the same ending. – farruhota Oct 27 '18 at 06:38
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That's another answer eligible to be accepted since it adresses what the OP is asking for. Maybe it would be better to do not give the full details for the solution but it is a minor issue. The method is very nice! – user Oct 27 '18 at 06:40
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Put $x =1/t$. Then $t$ tends to $0$.
$L= \frac{t-\ln(1+t)}{t^2}$.
$L= \frac{-t-\ln(1-t)}{t^2}$.
$2L = \frac{-\ln(1-t^2)}{t^2}=1$
Martin Sleziak
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maveric
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That's a nice trick but we need to assume that the limit exists and it is not a given of the question. – user Oct 27 '18 at 09:32
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even we when apply limits using l hospital rule , we assume it. then solve it. solving without expansions or series we need to assume existence and proceed. – maveric Oct 27 '18 at 15:23
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With l’Hopital rule we don’t need to assume that limit exists we show that $f’/g’ \to L \implies f/g \to L$, it seems a different thing to me. Isn’t it? – user Oct 27 '18 at 16:01