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This is the exercise:

$$f_{n}(x) = nx(1-x^2)^n,\qquad n \in {N}, f_{n}:[0,1] \to {R}.$$ Find ${f(x)=\lim\limits_{n\to\infty } {nx(1-x^2)^n}}$.

I know that $\forall x\in (0,1]$ $\Rightarrow (1-x^2) \in [0, 1) $ but I still don't know how to calculate the limit. $\lim\limits_{n\to\infty } {(1-x^2)^n}=0$ because $(1-x^2) \in [0, 1) $ and that means I have $\infty\cdot0$.

I tried transformation to $\frac{0}{0} $ and here is where I got stuck. I hope someone could help me.

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up vote 7 down vote accepted

Clearly it's enough to show that $\lim_{y \to \infty}yt^y=0$ when $t \in (0,1)$. Write this as $y/t^{-y}$. Applying L'Hopital's rule yields

$$\lim_{y \to \infty}yt^y=\lim_{y \to \infty}\frac{1}{-(\log t)t^{-y}}=0.$$

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Thank you for your help too. I wrote this in a notebook ! – NumLock Aug 10 '11 at 17:37

It is clear that if $x=0$ or $x=1$, then $f_n(x)=0$ for all $n$, so $\lim\limits_{n\to\infty}f_n(x) = 0$.

If $0\lt x\lt 1$, then $0\lt 1-x^2 \lt 1$. So the question amounts to asking why/whether $\lim\limits_{n\to\infty}nt^n=0$ if $0\lt t\lt 1$.

Note that if $0\lt t\lt 1$, then $\sum_{n=0}^{\infty}t^n$ converges to $\frac{1}{1-t}$. In fact, the sequence is the Taylor series for $\frac{1}{1-t}$ around $0$ on $(-1,1)$.

Since $\sum t^n$ is the Taylor series around $0$, within the interval of convergence we can differentiate term by term and obtain the Taylor series of the derivative. That is, $$\frac{1}{(1-t)^2} = \frac{d}{dt}\;\frac{1}{1-t} = \frac{d}{dt}\sum_{n=0}^{\infty}t^n = \sum_{n=0}^{\infty}\frac{d}{dt}t^n = \sum_{n=0}^{\infty}nt^{n-1} = \sum_{n=1}^{\infty}nt^{n-1}.$$ Since this series converges for any $t$ in $(-1,1)$, it follows that the sequence of terms coverges to $0$; that is, $$\lim_{n\to\infty} nt^{n-1} = 0\quad\text{for all }t\text{ with }|t|\lt 1.$$ Now notice that for $0\lt t\lt 1$ and $n\geq 1$, we have $$0 \leq (n-1)t^{n-1} \leq nt^{n-1}.$$ Applying the Squeeze Theorem, since both the constant sequence $0$ and the sequence $nt^{n-1}$ converge to $0$, it follows that the sequence $(n-1)t^{n-1}$ also converges to $0$ as $n\to\infty$. That is, $$\lim_{n\to\infty} (n-1)t^{n-1} = \lim_{n\to\infty}nt^n = 0\quad\text{if }0\lt t\lt 1.$$

Now putting it all together we have $$\lim_{n\to\infty} nx(1-x^2)^n = x\lim_{n\to\infty}n(1-x^2)^n = 0\quad\text{if }0\lt x\lt 1,$$ and the cases $x=0$ and $x=1$ are already taken care of.

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Thank you very much for your detalied explanation! – NumLock Aug 10 '11 at 17:35
Burno's answer is simpler, though. In my head, I did L'Hopital's Rule incorrectly and didn't double check, so I went the hard direction instead. – Arturo Magidin Aug 10 '11 at 17:36

$\lim\limits_{n\to\infty} nx(1-x^2)^n$ becomes $x \lim\limits_{n\to\infty} n(1-x^2)^n$ simply because $x$ is a factor that does not change as $n$ approaches $\infty$. Then you have $x \lim\limits_{n\to\infty} na^n$, and what you know about $a$ is that it's between 0 and 1. So it then becomes $x\lim\limits_{n\to\infty} \dfrac{n}{b^n}$ where $b=1/a$, so what you know about $b$ is that $b>1$.

It is easy to find that limit by using L'Hopital's rule, differentiating with respect to $n$. (Notice that $\dfrac{d}{dn} b^n$ is not $nb^{n-1}$ since we're differentiating with respect to $n$, not with respect to $b$. Rather $\dfrac{d}{dn} b^n = b^n\cdot \log_e b$.)

But although L'Hopital's rule gets you the bottom-line answer, it often doesn't give much insight. But now suppose there's a certain amount called $\Delta n$ by which $n$ has to increase to make $b^n$ three times as big. Since $n$ is in the exponent, every time the exponent increases by that same amount, $b^n$ becomes three times as big regardless of how big $n$ already was. So $b^{\Delta n} = 3$, or in other words, $\Delta n = \log_n 3$. So what happens to $\dfrac{n}{b^n}$ when $n$ increases by that much? The numerator hardly changes at all if $n$ is many times the size of $\Delta n$, but the denominator gets 3 times as big. So the fraction is very nearly cut down to $1/3$ of the size it was. Certainly it's less than $1/2$ the size it was if $n$ is big enough by comparison to $\Delta n$. So as $n$ approaches $\infty$, the fraction keeps repeatedly getting cut down to less than half what it was. That tells you what it approaches.

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Your first line is true only if $x\neq 0$. If $x=0$, then clearly $nx(1-x^2)^n=0$, so the limit exists and is equal to $0$; but $n(1-x^2)^n = n$ has no limit as $n\to \infty$, so $x\lim\limits_{n\to\infty}n(1-x^2)^n$ is not defined. – Arturo Magidin Aug 10 '11 at 17:55

Please see $\textbf{9.10}$ at this given link: The Solution is actually worked out for a more general case and I think that should help.

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Thank you very much! – NumLock Aug 10 '11 at 17:48
NUmlock: Welcom :) – user9413 Aug 10 '11 at 18:11

Three ways for showing that, for any $a \in (0,1)$ fixed, $$ \mathop {\lim }\limits_{n \to \infty } na^n = 0 $$ (which implies that $f(x)=0$).

1) Fix $b \in (a,1)$. Since $\lim _{n \to \infty } n^{1/n} = 1$, $n^{1/n}a < b$ $\forall n \geq N_0$. Hence, $na^n = (n^{1/n} a)^n < b^n $ $\forall n \geq N_0$; the result thus follows from $\lim _{n \to \infty } b^n = 0$.

2) If $b \in (a,1)$, then $$ \frac{{b - a}}{a}na^n = (b - a)na^{n - 1} = \int_a^b {na^{n - 1} \,dx} \le \int_a^b {nx^{n - 1} \,dx} = x^n |_a^b = b^n - a^n , $$ and so the result follows from $\lim _{n \to \infty } (b^n - a^n) = 0$.

3) Similarly to 2), if $b \in (a,1)$, then by the mean-value theorem $$ b^n - a^n = nc^{n - 1} (b - a) \ge na^{n - 1} (b - a) = na^n \frac{{b - a}}{a}, $$ for some $c \in (a,b)$; hence the result.

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