Find the value of the following limit: $$\huge\lim_{x\to\infty}e^{e^{e^{\biggl(x\,+\,e^{-\left(a+x+e^{\Large x}+e^{\Large e^x}\right)}\biggr)}}}-e^{e^{e^{x}}}$$

(original image)

I don't even know how to start with. (this problem was shared in Brilliant.org)

Some of the ideas I tried is to take the natural log of this expression and reduce it to $\ln(a/b)$ then use L'Hopital's but that made it false!!

I know the value of the limit it is $e^{-a}$ but please how to prove it?

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    $\begingroup$ That limit looks evil. $\endgroup$
    – homegrown
    Commented May 25, 2014 at 20:44
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    $\begingroup$ Not exactly a duplicate, but see also here. (The link points to an answer, not a question.) $\endgroup$ Commented May 25, 2014 at 21:54
  • $\begingroup$ Can you add the link to the solutions proposed on Brilliant.org (if there is any of course)? $\endgroup$
    – Dal
    Commented Dec 21, 2014 at 18:01
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    $\begingroup$ @AndrewD.Hwang page not found $\endgroup$
    – user153330
    Commented Mar 7, 2016 at 23:10
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    $\begingroup$ @user153330: Thanks. Apparently the linked question (a general request for "hard calculus problems that can be solved by a high school student") got deleted in August 2015 (along with all the answers, one of which was my suggestion of the limit in your question...). $\endgroup$ Commented Mar 7, 2016 at 23:45

4 Answers 4


Let's observe that the limit is of form $$ f\left(x+\frac{\alpha}{f'(x)}\right) - f\left(x\right) $$ where $f(x)=e^{e^{e^x}}$ and $\alpha=e^{-a}$. Since $f$ is differentiable, we have $$ f\left(x+\frac{\alpha}{f'(x)}\right) - f\left(x\right)= \frac{\alpha}{f'(x)}f'\left(x+\xi\right) $$ for some $0\leq \xi \leq \frac{\alpha}{f'(x)}$. It remains to prove that $f$ is "continuous enough" for this limit to converge as it seems it should, that is $$ \frac{f'\left(x+\xi\right)}{f'\left(x\right)}\to 1 $$ I'll try to find some clever way to do this.

Edit: Well, brutal force will do. We have $$ \xi\leq \alpha\, e^{-e^{e^x}}e^{-e^x}e^{-x} $$ Now, $$ \frac{f'\left(x+\xi\right)}{f'\left(x\right)}= e^{e^{e^{x+\xi}}-e^{e^x}} e^{e^{x+\xi}-e^x}e^\xi $$ Obviously, $e^\xi\to 1$. Furthermore, by above inequality, $$ e^{x+\xi}-e^x=e^x\left(e^\xi - 1\right)=e^xO\left(\xi\right)\to 0 $$ and $$ e^{e^{x+\xi}}-e^{e^x}=e^{e^x}\left(e^{e^{x+\xi} -e^x}-1\right)= e^{e^x}O\left(e^{x+\xi} -e^x\right)=e^{e^x}e^x O\left(\xi\right)\to 0 $$ That finishes the proof.


Here is a very simple way using only the asymptotic expansion $e^t \in 1+t+Θ(t^2)$ as $t \to 0$

As $x \to \infty$:

$\Large e^{e^{e^{x+e^{−\big(a+x+e^x+e^{e^x}\big)}}}}−e^{e^{e^x}}$

$\Large = e^{e^{e^xe^y}} − e^{e^{e^x}}$ where $\large y = e^{−e^{e^x}-e^x-x-a} \in o(1)$

$\Large \in e^{e^{e^x\big(1+y+Θ(y^2)\big)}}-e^{e^{e^x}} \subseteq e^{e^{e^x\Big(1+e^{−e^{e^x}-e^x-x-a}+Θ\big(e^{-2e^{e^x}-2e^x-2x}\big)\Big)}}-e^{e^{e^x}}$

$\Large = e^{e^{e^x}e^z}-e^{e^{e^x}}$ where $\large z = e^{-e^{e^x}-e^x-a}+Θ\big(e^{-2e^{e^x}-2e^x-x}\big) \subseteq o(1)$

$\Large \subseteq e^{e^{e^x}\big(1+z+Θ(z^2)\big)}-e^{e^{e^x}} \subseteq e^{e^{e^x}\Big(1+e^{−e^{e^x}-e^x-a}+Θ\big(e^{-2e^{e^x}-2e^x}\big)\Big)}-e^{e^{e^x}}$

$\Large = e^{e^{e^x}}e^w-e^{e^{e^x}}$ where $\large w = e^{−e^{e^x}-a}+Θ\big(e^{-2e^{e^x}-e^x}\big) \subseteq o(1)$

$\Large \subseteq e^{e^{e^x}} \large \big(1+w+Θ(w^2)\big) \Large - e^{e^{e^x}} \subseteq e^{e^{e^x}} \large \big(1+e^{−e^{e^x}-a}+Θ(e^{−2e^{e^x}})\big) \Large - e^{e^{e^x}}$

$\Large = e^{-a} + \large Θ\big(e^{−e^{e^x}}\big)$

Notice how fast the limit converges! If higher-order error terms are desired, one just has to increase the number of terms used in the expansion of the exponential function.


This method is completely rigorous (which can be verified by explicitly finding suitable values for the constants hidden in the asymptotic notation) and in fact this is roughly how the better computer algebra systems find generalized asymptotic expansions (even if there is no Taylor series). For the same reason, this method also does not require any thinking; just work from inside out and expand whatever can be expanded, simply keeping track of the error terms using asymptotic bounds.

One natural question is how to know how many terms are needed at each expansion. One way is to simply guess and see what happens. If at some point the significant term itself is cancelled out, we can then trace backward to the places where we need an asymptotically more precise expansion. If the final result has less terms than desired, we can likewise deterministically identify the places where we need more terms in the expansion.


It's been a while since I've had to compute such limits, but the general strategy is as follows:

Let $F(x) = \exp(\exp(\exp{x})))$ and $\delta(x) = \exp(-a - x - \exp{x} - \exp(\exp{x}))$. Then we are asked to compute the limit

$$ L = \lim_{x \rightarrow \infty} F(x + \delta(x)) - F(x) $$

Notice that $\delta(x) \rightarrow 0$. By differentiability, we can write $$ F(x + \delta(x)) = F(x) + F'(x) \delta(x) + E(x)$$ where $E(x)$ is an error term (very difficult to calculate explicitly). Then $$ L = \lim_{x \rightarrow \infty} F'(x) \delta(x) + E(x).$$ Notice that $F'(x) \delta(x) = \exp(-a)$ for all $x$. Therefore it remains to prove that $$ \lim_{x \rightarrow \infty} E(x) = 0$$ By Taylor's theorem, the remainder $E(x)$ satisfies the upper bound $$ |E(x)| \le \frac{ |f''(\xi)|}{2} \delta(x)^2, $$ for some $\xi$ such that $x \le \xi \le x + \delta(x)$. After explicitly calculating $f''(x + \delta)$, then one can see that this upper bound vanishes as $x \rightarrow \infty$.

  • $\begingroup$ That last limit is the difficult part. $\endgroup$
    – user153330
    Commented May 25, 2014 at 21:24
  • $\begingroup$ Yes, you're right =). I believe one can control $E(x)$ by using an upper bound on the Taylor series remainder, since $F(x)$ is in fact smooth. $\endgroup$ Commented May 25, 2014 at 21:27
  • $\begingroup$ I have updated my answer to reflect the strategy I mentioned above. $\endgroup$ Commented May 26, 2014 at 3:53
  • $\begingroup$ The definition of delta misses a () I can't edit because it's too little a change. $\endgroup$
    – chx
    Commented May 26, 2014 at 5:03

Let $ A=\exp\left(x+e^{-\left(a+x+e^x+e^{e^x}\right)}\right) $ and $B=\exp(x) $. Then we can easily conclude that $A/B$ tends to $1$, but a little more analysis allows us to infer that $A-B$ also tends to $0$.

We have $$A-B=B\cdot\frac{\exp(\log A-\log B) - 1}{\log A-\log B} \cdot(\log A-\log B) $$ and the middle fraction tends to $1$ and hence the limit of $A-B$ is same as that of $$B(\log A-\log B) $$ ie $$e^x\cdot e^{-\left(a+x+e^x+e^{e^x}\right)}$$ and thus $A-B$ tends to $0$.

It follows that $C/D\to 1$ where $C=e^A, D=e^B$. Applying same idea we note that limit of $C-D$ is same as that of $$D(\log C-\log D)=D(A-B) $$ which is same as that of $$DB(\log A-\log B) $$ ie $$e^{e^x} e^x\cdot e^{-\left(a+x+e^x+e^{e^x}\right)}$$ Thus we conclude that $C-D\to 0$.

Next let $E=e^C, F=e^D$ then $E/F\to 1$ and limit of $E-F$ is same as that of $$F(\log E-\log F) =F(C-D) $$ which is same as that of $$FDB(\log A-\log B) $$ ie $$e^{e^{e^x}} e^{e^x} e^x\cdot e^{-\left(a+x+e^x+e^{e^x}\right)}$$ The desired limit is then $e^{-a} $.


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