Limit of $n(1-x)^n$ as $n\to\infty$ when $0<x<1$

Question

Wolfram 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

Answer 1

Incrementing $n$ by $1$ has the effect of multiplying $n$ by $\dfrac{n+1} n,$ making it bigger, but also has the effect of multiplying $(1-x)^n$ by $(1-x),$ making it smaller. Which will prevail?

As $n$ grows, it ultimately gets immensely bigger than $\dfrac{1-x} x.$

When that happens, incrementing $n$ by $1$ has the effect of multiplying $n(1-x)^n$ by $\left( 1 + \dfrac 1 n \right)(1-x),$ which is a number less than $1$ and which gets smaller as $n$ grows.

So after reaching some large value of $n$, the rest of the sequence is bounded above by a geometric sequence with a common ratio less than $1;$ thus is approaches $0.$

Answer 2

We have

$$n(1-x)^n=e^{\log n+n\log (1-x)}\to 0$$

indeed

$$\log n+n\log (1-x)=n\left(\frac{\log n}n+\log (1-x)\right)\to -\infty$$

since $\frac{\log n}n\to 0$ and $\log (1-x)<0$.

Answer 3

I mean, if you really want to use L'Hôpital, you can write it as $$ \frac{n}{(1-x)^{-n}}, $$ and then differentiating numerator and denominator gives $$ \frac{1}{-n(1-x)^{-n-1}} = \frac{-1}{1-x} \frac{(1-x)^n}{n}, $$ and the right hand side tends to $0$ since both variable terms in the product do.

Answer 4

Take the series $\;\sum\limits_{n=1}^\infty n (1-x)^n\;$ and apply the $\;n\,-$ th root test to its absolute value:

$$\sqrt[n]{n|1-x|^n}=\sqrt[n]n\,|1-x|\xrightarrow[n\to\infty]{}|1-x|$$

and we thus get absolute convergence iff

$$|1-x|<1\implies -1<1-x<1\implies -2<-x<0\implies 0<x<2$$

so the series' sequence's general term tends to zero.

Answer 5

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.