# Proof that $\lim\limits_{x\to \infty} (1+\frac{1}{\ln x})^x = \infty$

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.

• 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. – saulspatz Mar 9 at 17:49

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

• 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. – Luca Ricchi Mar 9 at 17:43
• One can use L'Hôpitals rule to prove this limit is infinite. – Peter Foreman Mar 9 at 17:44
• Can you accept my solution? – Peter Foreman Mar 9 at 18:03
• 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 – Luca Ricchi Mar 9 at 18:05

Taking $$\log$$ :

$$f(x):=$$

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

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

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 $$x \rightarrow \infty.$$.