Find $\lim_{n \to \infty}f_n(x)$

Question

Find $$\lim_{n \to \infty}f_n(x)$$ where $$f_n(x)=n^2x(1-x)^n$$ $0 \lt x \lt 1$

My try:

By symmetry $$\lim_{n \to \infty}f_n(x)=\lim_{n \to \infty}f_n(1-x)=\lim_{n \to \infty}n^2(1-x)x^n=(1-x)\lim_{n \to \infty}n^2 x^n$$

Now $$\lim_{n \to \infty}n^2 x^n=\lim_{n \to \infty}\frac{x^n}{\frac{1}{n^2}}$$

Now can we use L'hopital's rule here?

Answer 1

Hint: Use L'hopital rule to find $\lim_{n \to \infty} f_n(x)$, where $$f_n(x) = x \frac{n^2}{(1-x)^{-n}}.$$

Answer 2

Differentiate to find the max. If it goes to zero, that's the limit; otherwise the limit depends on $x$.

If $f_n(x)=n^2x(1-x)^n$, then

$\begin{array}\\ f_n'(x) &=n^2(x(1-x)^n)'\\ &=n^2((1-x)^n-nx(1-x)^{n-1})\\ &=n^2(1-x)^{n-1}((1-x)-nx)\\ &=n^2(1-x)^{n-1}(1-(n+1)x)\\ &=0 \qquad\text{for } x=1/(n+1) \end{array} $

At $x=1/(n+1)$,

$\begin{array}\\ f_n(x) &=n^2(1/(n+1))(1-1/(n+1))^n\\ &=n^2(1/(n+1))(n/(n+1))^n\\ &=n\dfrac1{(1+1/n)(1+1/n)^n}\\ &\approx n\dfrac1{e}\\ &\to \infty\\ \end{array} $

So $\lim_{n \to \infty} f_n(0)$ does not exist since $f_n(0) = 0$.

If $f_n(x) =x(1-x)^n$ then $\lim_{n \to \infty} f_n(x) = 0$ for $0 \le x \le 1$.

Answer 3

An option:

For $0< x <1$, show that

$\lim_{n \rightarrow \infty} n^2x^n =0.$

Set $x=e^{-y} , y>0$, and consider $\dfrac{n^2}{e^{ny}}$.

$e^{ny} =$

$ 1+ ny +(ny)^2/2! + (ny)^3/3! +.. \gt (ny)^3/3!$.

Hence :

$\dfrac{n^2}{e^{ny}} \lt \dfrac{(3!)n^2}{n^3y^3}= (\dfrac{3!}{y^3})(\dfrac{1}{n}).$

The limit $n \rightarrow \infty$ is?

Answer 4

Denote $y=\dfrac{1}{x}$. Since $0<x<1$,then $y>1.$ Hence $$\lim_{n \to \infty}n^2 x^n=\lim_{n \to \infty}\frac{n^2}{y^n}=\lim_{n \to \infty}\frac{2n}{y^n\ln y}=\lim_{n \to \infty}\frac{2}{y^n(\ln y)^2}=0.$$

Answer 5

It might be easier to use the Ratio Test: $$ \begin{align} \lim_{n\to\infty}\frac{f_{n+1}(x)}{f_n(x)} &=\lim_{n\to\infty}\frac{(n+1)^2}{n^2}(1-x)\\ &=1-x \end{align} $$ For $0\lt x\lt1$, we have $0\lt1-x\lt1$. Therefore, the series $$ \sum_{n=0}^\infty f_n(x) $$ converges. Thus, the terms go to zero. That is, $$ \lim_{n\to\infty}f_n(x)=0 $$