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How can I find this limit: $$\lim_{n \to +\infty} n^2x(1-x)^n, x \in [0,1]$$

Do I have to use the L'Hospital rule? If so, do I have to differentiate with respect to $n$ or to $x$ ?

EDIT:

I also tried to find the supremum of $n^2x(1-x)^n$..I found that $f(x)=n^2x(1-x)^n$ achieves its maximum at the point $\frac{1}{n+1}$ and that $\sup_{x \in [0,1]} n^2x(1-x)^n=\frac{n}{e} \to +\infty$

Could you tell me if it is right?

Simon S
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evinda
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    With respect to $n$. Note $x=0$ or $x=1$ are trivial cases, and don't require anything fancy. – David Mitra Jun 25 '14 at 12:29
  • Since $1-x^2<0,\forall x\in(0,1)$;

    $\lim\limits_{n\to\infty}n^2x(1-x)^n=\lim\limits_{n\to 0} n^2\exp(n\ln(1-x^2))=0$, $\forall x\in(0,1)$

    for $x\in{0,1}$ it should be clear.

     – OBDA Jun 25 '14 at 12:40
  • You can factor out $x$ and note that $1-x$ is a positive constant $<1$ (trivial limit cases $0$ and $1$ handled separately). Your limit is a product of a power by a negative exponential, $n^2e^{-pn}$. The latter are known to decrease faster than any power. –  Jun 25 '14 at 12:40
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    Your derivation of the maximum is correct. It simply shows that the convergence is not uniform. – Frédéric Jun 25 '14 at 13:05
  • @Frédéric Nice,thank you very much!!!! – evinda Jun 25 '14 at 13:07
  • Hmm... if I fill in $x=\frac 1{n+1}$ I do not get $\frac n e$... – Klaas van Aarsen Jun 25 '14 at 20:49

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In addition to the trivial cases pointed out by David Mitra, you can `pull out' $x$ out of the limit -it's a constant and look at $Lf = \log f(x)$. $$ Lf = 2 \log x +n \log(1-x) $$ Clearly the expression is $O(n)$, and $\log(1-x)<0$, so $Lf \to -\infty.$ Exponentiating it back, you get the limit of $0$.

Alex
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This limit is always zero, there are two cases:

1)$x=0$-it's obvious.

2)$x \neq 0$, then the limit is zero, because $1-x <1$ and exponential function decrease faster than polynomial $n^2$

agha
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You do not have to apply L'hospital.

Let $x \neq 0$ or $1$ (trivial cases). Define $x_n=n^{2}x(1-x)^n$. We have

$$ x_{n+1}=\frac{(n+1)^2}{n^2}(1-x)x_n $$

implying ($x\in $(0,1)) that there is a value of $n$ after which the $x_{n+1}<x_n$. Furthermore $x_n>0$, hence it converges to $L$. Taking the limit on both sides of the equation we get

$$ L=(1-x)L $$ leading to $L=0$.

Frédéric
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