I have been trying to prove that $$ \lim\limits_{x\to \infty} (1+\frac{1}{\ln x})^x = \infty $$ and this is what I got: $$ \lim\limits_{x\to \infty} (1+\frac{1}{\ln x})^x = \lim\limits_{x\to \infty} e^{\ln (1+\frac{1}{\ln x})^x} = \lim\limits_{x\to \infty} e^{x * \ln (1+\frac{1}{\ln x})} $$ Then due to the fact that the e function is continuous and that $a*b=\frac{a}{\frac1b}$ $$ = e^{ \lim\limits_{x\to \infty} \frac{\ln (1+\frac{1}{\ln x})}{\frac1x} } $$ Since both the top and the bottom go to 0 as ${x\to \infty}$ we can apply L'Hospital and after deriving both we get $$ e^{ \lim\limits_{x\to \infty} \frac{-\frac{1}{x*\ln x+x*\ln^2 x}}{-\frac{1}{x^2}}} = e^{ \lim\limits_{x\to \infty} \frac{x^2}{x*(\ln x+\ln^2 x)}} = e^{ \lim\limits_{x\to \infty} \frac{x}{\ln x+\ln^2 x}} $$ and then since the x function grows much more rapidly than the logarithmic functions at any power, we get $$ e^{ \lim\limits_{x\to \infty} \frac{x}{\ln x+\ln^2 x}} = e^{\infty} = \infty $$ Since I am quite new to calculus I don't feel sure at all about what I just did so it would be great if I could get some feedback from experienced people. Also it's my first post here, I hope I didn't break any rule. In the meantime, I wish everyone a nice day.

  • $\begingroup$ This is correct, so far as I can see, and the post is fine. +1 I would suggest writing $f(x) = \ln{(1+1/\ln{x})}^x$ and evaluating $\lim_{x\to\infty}f(x),$ though. That way you get rid of all those exponentials, and it easier to read. Once you find that the limit is $\infty$ you know the limit you seek is $e^\infty.$ The difference from what you did is mainly typographical. $\endgroup$ – saulspatz Mar 9 at 17:49

This limit is equal to $$\lim_{x\to\infty} \bigg(\big(1+\frac{1}{\ln{x}}\big)^{\ln{x}}\bigg)^{(\frac{x}{\ln{x}})}=e^{\lim_{x\to\infty}\frac{x}{\ln{x}}}=e^{\infty}=\infty$$

  • $\begingroup$ Thank you very much. So the fact the logarithmic function grows slowlier than the simple x is generally accepted and does not need particular other proofs, I assume. $\endgroup$ – Luca Ricchi Mar 9 at 17:43
  • $\begingroup$ One can use L'Hôpitals rule to prove this limit is infinite. $\endgroup$ – Peter Foreman Mar 9 at 17:44
  • $\begingroup$ Can you accept my solution? $\endgroup$ – Peter Foreman Mar 9 at 18:03
  • $\begingroup$ of course, I find it very elegant, thanks again but I can't upvote because I am new and I still don't have 15 points of reputation $\endgroup$ – Luca Ricchi Mar 9 at 18:05

Taking $\log$ :


$x (\log (1+\dfrac{1}{\log x})-\log 1)=$

$(\dfrac{x}{\log x})\dfrac{\log (1+\dfrac{1}{\log x})-1}{\dfrac{1}{\log x}}.$

Set $h:=\dfrac{1}{\log x}$:

$x \rightarrow \infty$ implies $h \rightarrow 0^+$.

$\lim_{ h \rightarrow 0^+}\dfrac{\log (1+h)-1}{h}=$

$ \log '(1)=1.$

Hence for small enough $h$:

$\dfrac{\log (1+h)-1}{h} >1/2.$

For large enough $x$:

$(1/2)\dfrac{x}{\log x}< f(x)$, i.e not bounded above.

Exponentiate and take the limit $x \rightarrow \infty.$.


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