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Let's consider the function $f:\mathbb{R}\rightarrow(0,\infty)$, with $f(x)\cdot \ln f(x)=e^x$, $\forall x \in \mathbb{R}$. Then compute

$$\lim_{x\to\infty}\left(1+\frac{\ln x}{f(x)}\right)^{\displaystyle\frac{f(x)}{x}}$$

The first solution

Since $$f(x)\cdot \ln f(x)=e^x, \forall x \in \mathbb{R}$$ we may easily deduce that $$\lim_{x\to\infty}f(x)=\infty$$ On the other hand $$f^2(x)> f(x)\cdot \ln f(x)=e^x$$ $$f(x)> e^{x/2}$$ and then $$0\le\lim_{x\to\infty} \frac{\ln x}{f(x)}\le\frac{\ln x}{e^{x/2}}\rightarrow 0$$ $$\lim_{x\to\infty} \frac{\ln x}{f(x)}=0$$ $$\lim_{x\to\infty} \displaystyle\frac{e^{x/2}}{x} \le \lim_{x\to\infty} \displaystyle\frac{f(x)}{x}=\infty$$

At this point we recognize that our limit case is $1^{\infty}$, and have $$\lim_{x\to\infty}\left(1+\frac{\ln x}{f(x)}\right)^{\displaystyle\frac{f(x)}{x}}=\lim_{x\to\infty}e^{\displaystyle \frac{\ln x}{x }}=e^0=1$$

The second solution (from a brilliant friend of mine - - so sorry I missed this way)

Let's take log of both sides of the limit

$$\ln L = \lim_{x\to\infty} \frac{f(x)}{\ln x} \ln \left(1+\frac{\ln x}{f(x)}\right)\times \lim_{x\to\infty} \frac{\ln x}{x}=1\times 0=0$$ that is simply justified by the fact that $f(x)>>x$ (see $f(x)> e^{x/2}$)

Question: how would you approach this question? Thank you.

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Thanks for sharing us bunch of interesting problems, sis. +1 – Babak S. Mar 2 '13 at 16:49
@BabakS.: I'm glad you like them! :-) Thanks. – user 1618033 Mar 2 '13 at 16:51
up vote 4 down vote accepted

This is a very good point knowing that if $$\lim_{x\to{+\infty}} f(x)^{g(x)}=1^{+\infty}$$ which is indeterminate limit then we can solve it by taking the following limit: $$k =\lim_{x\to +\infty}\big(f(x)-1\big)g(x)$$ instead. So $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^k$$

Now, try to use this formula also. It gives you $~~\text{e}~~$ at last. For more see how @Brian proved me that. This proof deserves more that +100. 100

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Thanks for your work. (+1) – user 1618033 Mar 2 '13 at 12:36
Good day, Babak! :+) – amWhy Mar 2 '13 at 13:58

$\lim\limits_{x\to+\infty}\left(1+\frac{\ln x}{f(x)}\right)^{\displaystyle\frac{f(x)}{x}}$

$\left(1+\frac{x}{f(e^x)}\right)^{\displaystyle\frac{f(e^x)}{e^x}} = \left(\left(1+\frac{x}{f(e^x)}\right)^{\displaystyle f(e^x)}\right)^{\displaystyle\frac{1}{e^x}} \sim\big(e^x\big)^{\displaystyle\frac{1}{e^x}} = \big(e\big)^{\displaystyle\frac{x}{e^x}} \to e^0 = 1$

The substitution $x \leftarrow e^x$ is justified by the fact that $x \to +\infty \Leftrightarrow e^x \to +\infty$

The $\sim$ by the fact that $f(e^x) \to +\infty$ (not completely sure about that)

And the rest is simple.

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Why substitute $\log x = t$ (is this what you meant to do?) when the rest of the manipulation can be done without it? – Alex Feb 11 '13 at 2:02
@Alex: How do you get $\left(1+\frac{x}{y}\right)^y$ then? – xavierm02 Feb 11 '13 at 2:12
If $f(x) \to \infty$, then $(1+\frac{\log x}{f(x)})^{f(x)}$ and $(1+\frac{t}{f(e^t)})^{f(e(t)}$ are the same, although I'm not sure at all that this is approximation using the definition of $e$ is correct at all, as the OP's expression involves $x$ both in the numerator ($\log x$) and the denominator of the power. – Alex Feb 11 '13 at 2:42
Thanks for your work. (+1) – user 1618033 Mar 2 '13 at 12:37

The first step is to notice that $f$ is differentiable: Differentiating the equation for $f$ with respect to $f$ (we're going to use the inverse function theorem here, so we're thinking of $x$ as a function of $f$) and dividing through by $e^x$ we get $$\frac{dx}{df}= \frac{1+\ln{f}}{e^x}=\frac{1+\frac{e^x}{f}}{e^x}.$$ But $f$ is by definition always positive. Therefore, the $dx/df$ can never be zero, which by the inverse function theorem means that $f$ must always be differentiable. Moreover, since $dx/df$ is always positive, so is $df/dx$, which means--and here's the punch line--that $f$ is (strictly) monotone increasing.

So, now let's try and find some bounds on $f$. Since $f$ is monotone increasing, it's going to eventually be strictly positive (in fact it one can show that it'll always be positive, but that's not going to be important to us), so we can use the fact that $\ln{x}<x$ for all $x>0$ to get that $\ln{f(x)}<f(x)$ for large enough $x$. Therefore, returning to the defining equation for $f$, we get the two inequalities $$e^x=f(x)\ln{f(x)}\leq f(x)^2$$ and $$e^x\geq[\ln{f(x)}]^2.$$ These simplify to $$e^{x/2}\leq f(x) \leq e^{e^{x/2}}.$$

Finally, let's head over to our limit, the inside of which I'll call $y$. Taking the natural log, we get $$\ln{y} = \frac{f(x)}{x}\ln\left(1+\frac{\ln{x}}{f(x)}\right).$$ Using $f(x)\geq e^{x/2}$, we can get rid of the $f(x)$ outside of the $\ln$, and using $f(x)\leq e^{e^{x/2}}$, we can get rid of the one inside. This gives $$\ln{y} \leq \frac{e^{x/2}}{x} \ln\left(1+e^{-e^{x/2}}\log{x}\right),$$ which I'll leave to you to show approaches $0$ as $x\to \infty$.

We'd like to simply stop here and say that $\ln{y}$ must also approach $0$, but this only works if $\ln{y}$ is (eventually) nonnegative---then we can use the squeeze theorem to conclude that yes, $\ln{y}$ does in fact approach $0$. So, is $\ln{y}$ nonnegative? Well, let's look back at the original equation for $\ln{y}$. Since $f$ is always positive, the part outside the $\ln$ is also positive (as long as $x>0$, but we're working with large $x$ here, so we can assume this). The same reasoning tells us that the part inside the $\ln$ must be greater than $1$ (as long as $x>1$, but again, we can assume this), and hence that the $\ln$ part itself is positive. Thus, $\ln{y}$ is indeed positive, and we get that $\ln{y}\to 0$ as $x\to 0$.

Hence, $$\lim_{x\to\infty} \left(1+\frac{\ln{x}}{f(x)}\right)^{\dfrac{f(x)}{x}}=1.$$

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Thanks for your work. (+1) – user 1618033 Mar 2 '13 at 12:35

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