Limit of $n(1-x)^n$ as $n\to\infty$ when $0Wolfram alpha gives that this is $0$, but I'm not sure how to show it.
I Tried writing as $\frac{(1-x)^n}{1/n}$ and using L'Hopital's rule, but a new $n$ term shows up every time I take the derivative.
I also try to set $h=\frac1n$ and then write as
$$
\lim_{n\to\infty} n(1-x)^n = \lim_{h\to 0} \frac{(1-x)^\frac1h}{h}
$$
but same problem, numerator goes to $0$, denominator to $0$, and L'Hôpital doesn't seem to work because the derivative of the numerator gives $\frac1h (1-x)^{\frac1h -1}$, which has the initial trouble

You can solve this by taking the logarithm of each side. Let $y = \lim_{n \rightarrow \infty} n(1-x)^n$, and then we see that $$\ln(y) = \ln(\lim_{n \rightarrow \infty} n(1 -x)^n) = \lim_{n\rightarrow \infty } \ln(n (1-x)^n)$$ by using the continuity of the natural log function. Now, we use log rules and see $$\ln(y) = \lim_{n \rightarrow \infty} \ln(n) + n\ln(1-x).$$ Since $1 - x$ is less than 1, $\ln(1-x) < 0$. The $\ln(n)$ term grows logarithmically in $n$ and the $n\ln(1-x)$ term grows linearly in $n$. So $\ln(y)$ tends to $- \infty$, hence $y = e^{\ln(y)}$ tends to 0.
