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

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

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?
• $$(1-\log x/x)^{x/\log x}\to e^{-1}$$ so $(1-\log x/x)^x\to 0.$ May 13, 2022 at 20:57

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

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!

• Hi, my friend ! May 14, 2022 at 3:23
• Hi Claude! How are you my friend? May 14, 2022 at 14:00

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 $$=0$$.

• 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?'
– Mike
May 13, 2022 at 22:12
• @Mike When we use $\sim$, the RHS is never zero. May 13, 2022 at 22:49

$$y=\left(1-\frac{\log (x)}{x}\right)^x\implies \log(y)=x \log \left(1-\frac{\log (x)}{x}\right)$$ 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 $$0.008963627$$