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

The answer is 0 (I have faith in my lecturer, so I believe this to be correct), but I get 1. I applied L'Hopital to the fraction, got $\lim_{x \to \infty} \frac{1}{x}$, and eventually $1$.

Questions:

  1. How do I reach $0$?
  2. I may agree that for $x \to 0$ there may be issues, but for $x \to \infty$ the function is well behaved (i.e. continuous): then why can't I calculate the limes inside? In other words, why does the approach above fail?
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    $\begingroup$ $$(1-\log x/x)^{x/\log x}\to e^{-1}$$ so $(1-\log x/x)^x\to 0.$ $\endgroup$ May 13, 2022 at 20:57

4 Answers 4

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Since $\frac x{\log x}\to \infty$ as $x\to \infty$ and $\left(1-\frac1y\right)^y\to e^{-1}$ as $y\to\infty,$ we have: $$\lim_{x\to\infty}\left(1-\frac{\log x}x\right)^{x/\log x}=e^{-1}$$

This means, in particular, for large $x,$ $$0<\left(1-\frac{\log x}x\right)^{x/\log x}<\frac12$$

So, for large $x,$ $$0<\left(1-\frac{\log x}x\right)^x<\left(\frac12\right)^{\log x}=\frac1{2^{\log x}}=\frac1{x^{\log 2}}\to 0.$$

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Using the inequality $\log(1+x)\le x$ for all $x>0$, we have

$$\begin{align}\left|\left(1-\frac{\log(x)}{x}\right)^x\right|&=e^{x\log(\left(1-\log(x)/x)\right)}\\\\ &\le e^{-\log(x)}\\\\ &=\frac1x \end{align}$$

whence applying the squeeze theorem yields the coveted result

$$\lim_{x\to\infty}\left(1-\frac{\log(x)}{x}\right)^x=0$$

as was to be shown!

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  • $\begingroup$ Hi, my friend ! $\endgroup$ May 14, 2022 at 3:23
  • $\begingroup$ Hi Claude! How are you my friend? $\endgroup$
    – Mark Viola
    May 14, 2022 at 14:00
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If we take logarithm, we will compute

$$\lim_{x\to+\infty}x\ln(1-\frac{\ln(x)}{x})$$

but $$\lim_{x\to +\infty}\frac{\ln(x)}{x}=0$$

and we know that

$$\ln(1+X)\sim X \;\;(X\to 0)$$

so, $$\ln(1-\frac{\ln(x)}{x})\sim -\frac{\ln(x)}{x}\;\,(x\to+\infty)$$

we just need to compute $$\lim_{x\to×\infty}x(-\frac{\ln(x)}{x})=-\infty$$

thus, your limit is zero $=0$.

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  • $\begingroup$ Where I could see a student getting tripped up though is in looking at this and wondering 'why wouldn't $\ln\big(1-\frac{\ln x}{x}\big) \approx 0$ be valid?' $\endgroup$
    – Mike
    May 13, 2022 at 22:12
  • $\begingroup$ @Mike When we use $ \sim $, the RHS is never zero. $\endgroup$ May 13, 2022 at 22:49
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Just for your curiosity.

You could easily go much beyond the limit.

$$y=\left(1-\frac{\log (x)}{x}\right)^x\implies \log(y)=x \log \left(1-\frac{\log (x)}{x}\right)$$ Using the Taylor expansion of $\log(1-t)$ when $t$ is small and replacing $t$ by $\frac{\log (x)}{x}$,we have $$\log(y)=x\Bigg[-\frac{\log (x)}{x}-\frac{\log ^2(x)}{2 x^2}-\frac{\log ^3(x)}{3 x^3}+\cdots \Bigg]$$ $$\log(y)=-\log (x)-\frac{\log ^2(x)}{2 x}-\frac{\log ^3(x)}{3 x^2}+\cdots$$ Now, using $y=e^{\log(y)}$ and Taylor again $$y=e^{\log(y)}=\frac{1}{x}-\frac{\log ^2(x)}{2 x^2}+\frac{\log ^3(x) (3 \log (x)-8)}{24 x^3}+\cdots$$

Try the above with $x=100$ (quite far away from infinity). The decimal representation of the result is $0.008963286$ while the exact value is $0.008963627$

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