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How to find the limit $$\lim_{x\to+\infty}\left(x-x^2\log\left(1+\frac{1}{x}\right)\right)$$ in a elementary way? I can solve with Taylor expansion, but it is placed in the beginning of my calculus book, so I should only use things like:

-Main theorems involving limits, including the limits for $x\to 0$, $\lim\frac{\sin x}{x}$, $\lim\frac{e^x-1}{x}$, $\lim\frac{\log(x+1)}{x}$, $\lim\frac{(x+1)^p-1}{x}$

-$\frac{x}{1+x} \leq \log(1+x) \leq x$ or similar inequalities

I cannot use derivatives, Taylor expansion, $o(x), O(x)$ and similar things.

Using the inequality that I have written above and the substitution $x=\frac{1}{\sin t}$ I have been only able to prove that the limit is greater or equal than 0 and smaller or equal than 1.

Any ideas?

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  • $\begingroup$ The first term goes to $+\infty$ and the second is positive for positive $x$. $\endgroup$ – Travis Aug 30 '14 at 16:57
  • $\begingroup$ Sorry, now the problem is correct. $\endgroup$ – Nisba Aug 30 '14 at 17:01
  • $\begingroup$ this limit can not be obtained via the results you have given in the question. You need more results (like further inequalities in my answer) to solve this. However using the results mentioned in the question one can show that "if this limit exist then it must be $1/2$", but the existence of this limit needs more analysis. $\endgroup$ – Paramanand Singh Aug 31 '14 at 7:02
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Let $t = 1/x$ so that $t \to 0^{+}$ as $x \to \infty$. Then the desired limit is $\lim\limits_{t \to 0^{+}}\dfrac{t - \log(1 + t)}{t^{2}}$. If we have $y > 0$ then we know that $1 - y^{2} < 1 < 1 + y^{3}$ and dividing by $(1 + y)$ we get $$1 - y < \dfrac{1}{1 + y} < 1 - y + y^{2}$$ Integrating (with respect to $y$) in the interval $[0, t]$ we get $$t - \frac{t^{2}}{2} < \log(1 + t) < t - \frac{t^{2}}{2} + \frac{t^{3}}{3}$$ for $t > 0$. this means that $$ \frac{1}{2} - \frac{t}{3} < \frac{t - \log(1 + t)}{t^{2}} < \frac{1}{2}$$ for $t > 0$. Now using squeeze theorem we get the desired limit as $1/2$. Note that I have used integration to establish the logarithmic inequalities. The inequalities can be established by using any other definition of $\log x$ also, but the definition based on integrals seems to be the easiest to deal with.

Update: Based on OP comment I think it is better to transform the limit into one involving exponential functions. Let $1 + t = e^{z}$ so that $ t \to 0^{+}$ implies $z \to 0^{+}$. Then we have $$\begin{aligned}L &= \lim_{t \to 0^{+}}\frac{t - \log(1 + t)}{t^{2}}\\ &= \lim_{z \to 0^{+}}\frac{e^{z} - 1 - z}{(e^{z} - 1)^{2}}\\ &= \lim_{z \to 0^{+}}\frac{e^{z} - 1 - z}{z^{2}}\cdot\left(\frac{z}{e^{z} - 1}\right)^{2}\\ &= \lim_{z \to 0^{+}}\frac{e^{z} - 1 - z}{z^{2}}\end{aligned}$$ Now there are two ways. First the easier one as follows. We have $$\begin{aligned}\frac{e^{z} - 1 - z}{z^{2}} &= \frac{1}{2!} + \frac{z}{3!} + \frac{z^{2}}{4!} + \cdots\\ &= \frac{1}{2} + \phi(z)\end{aligned}$$ where $$\phi(z) = \frac{z}{3!} + \frac{z^{2}}{4!} + \cdots$$ so that $$0 < \phi(z) \leq \frac{z}{6} + \frac{z^{2}}{18} + \frac{z^{3}}{54} + \cdots$$ i.e $$0 < \phi(z) \leq \frac{z/6}{1 - (z/3)} = \frac{z}{6 - 2z}$$ for $0 < z < 3$. By Squeeze theorem as $z \to 0^{+}$ we have $\phi(z) \to 0$. Hence $(e^{z} - 1 - z)/z^{2} \to 1/2$.

The hard part is to show that $(e^{z} - 1 - z)/z^{2} \to 1/2$ without using the series for $e^{z}$ and instead using the definition $e^{z} = \lim_{n \to \infty}(1 + z/n)^{n}$ directly. Let me know if you are interested in the hard version.

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  • $\begingroup$ Thank you! My book defines the logarithm as the inverse of $e^x$. $\endgroup$ – Nisba Aug 31 '14 at 8:25
  • $\begingroup$ @LucaMarconato: And how does it define $e^{x}$? Then I may be able to explain better. $\endgroup$ – Paramanand Singh Aug 31 '14 at 8:27
  • $\begingroup$ $\lim(1+\frac{x}{n})^n$, and it is proved also the representation $\sum\frac{x^k}{k!}$. $\endgroup$ – Nisba Aug 31 '14 at 8:29
  • $\begingroup$ @LucaMarconato: I will update my answer based on this information provided by you. $\endgroup$ – Paramanand Singh Aug 31 '14 at 8:32
  • $\begingroup$ Thank you and I will try to get those inequalities. $\endgroup$ – Nisba Aug 31 '14 at 8:45
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Here is how. We have

$$ L = \lim_{x\to \infty} x^2\left( \frac{1}{x} - \ln\left(1+\frac{1}{x}\right) \right). $$

Letting $y=\frac{1}{x}$ results in the form

$$ L = \lim_{y\to 0} \frac{1}{y^2}(y-\ln(1+y)) = -\lim_{y\to 0 } \frac{\frac{\ln(1+y)}{y}-1}{y-0}= -\lim_{y\to 0 } \frac{f(y)-1}{y-0}, $$

where

$$ f(y) = \frac{\ln(1+y)}{y}. $$

I leave it here for you.

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  • $\begingroup$ I would very much like to see how one can continue from here. $\endgroup$ – Quang Hoang Aug 30 '14 at 19:09
  • $\begingroup$ @QuangHoang: This matches the form $\lim\frac{e^x-1}{x}.$ $\endgroup$ – Mhenni Benghorbal Aug 30 '14 at 19:14
  • $\begingroup$ Maybe I really need a sleep. Will return later to see how it evolves. For now I can only see that this is equivalent to the second derivative of $e^x$. $\endgroup$ – Quang Hoang Aug 30 '14 at 19:22
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    $\begingroup$ i don't think we can continue from here with the results mentioned in the question. we need more tools to deal with it. $\endgroup$ – Paramanand Singh Aug 31 '14 at 7:04
  • $\begingroup$ This is circular. If $f(y) = \frac{\log (1+y)}{y}$, we can differentiate this using standard techniques only for $y \neq 0$. At $y=0$, we have to the evaluate the limit explicitly. Therefore, this method leads nowhere. $\endgroup$ – MathematicsStudent1122 Nov 12 '16 at 5:04
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Hint: $$x\log \bigg(1+\frac{1}{x}\bigg)=\log \bigg(1+\frac{1}{x}\bigg)^x$$

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  • $\begingroup$ Already tried, one more hint? $\endgroup$ – Nisba Aug 30 '14 at 17:06
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    $\begingroup$ The result is $\frac{1}{2}$, your suggestion leads me to an indeterminate form "$(+\infty) - (+\infty)$". $\endgroup$ – Nisba Aug 30 '14 at 17:14

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