Limitless potential

$\begin{aligned} L &= \lim\limits_{n \to 0^{+}}\left(\frac{1}{n}\right)^n \\ &= \lim\limits_{n \to 0^{+}}e^{n\ln\left(\frac{1}{n}\right)} \\ &= e^{\blue{\lim\limits_{n \to 0^{+}} n \ln\left(\frac{1}{n}\right)}} \end{aligned}$ Now, we just need to find the limit in the exponent: $\begin{aligned} \blue{\lim\limits_{n \to 0^{+}} n \ln\left(\frac{1}{n}\right)} &= \lim\limits_{n \to 0^{+}} n\ln(1) - n\ln(n) \\ &= \lim\limits_{n \to 0^{+}} - n\ln(n) \\ &= \lim\limits_{n \to 0^{+}} \frac{\ln(n)}{\frac{-1}{n}} \\ \end{aligned}$ Since plugging in $0$ here gives us a $\cfrac{\infty}{\infty}$ indeterminate form, we can use L'Hopital's Rule and differentiate the numerator and the denominator: $\begin{aligned} \implies \lim\limits_{n \to 0^{+}} \frac{\frac{d}{dx}(\ln(n))}{\frac{d}{dx}\left(\frac{-1}{n}\right)} &= \lim\limits_{n \to 0^{+}} \frac{\frac{1}{n}}{\frac{1}{n^2}} \\ &= \lim\limits_{n \to 0^{+}} n \\ &= \orange{\boxed{0}} \end{aligned}$

Now we can plug this in as our exponent: $L = e^{\orange{0}} = \boxed{1}$

Since we need $\lfloor 1000 L \rfloor$ as our answer, we plug in $1$ as $L$ to see that $\lfloor 1000\blue{L} \rfloor = \lfloor 1000\blue{(1)} \rfloor = \green{\boxed{1000}}$