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:

*

*How do I reach $0$?

*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?

 A: 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 $x,$ $$0<\left(1-\frac{\log x}x\right)^x<\left(\frac12\right)^{\log x}=\frac1{2^{\log x}}=\frac1{x^{\log 2}}\to 0.$$
A: 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!
A: 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 $=0$.
A: 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     $0.008963627$
