Nice exercise based on IVT theorem and its application

Exercise:

Consider the function ${{f}_{n}}\left( x \right)={{x}^{n}}-nx+1\,\,,\,\,n\in \mathbb{N}$ such that $n\ge 3$

1) Show that , ${{f}_{n}}\left( x \right)=0$ has unique solution ${{\alpha }_{n}}\in \left[ 0,1 \right]$

2) Compare ${{\alpha }_{n}}\,,\,\frac{1}{n}\,\,\And \,\,\frac{2}{n}$

3) Find the limit of ${{\alpha }_{n}}$ and $n{{\alpha }_{n}}$ as $n\to \infty $

Solution:

1) We have${{f}_{n}}$ is a continuous function in$\mathbb{R}$ (${{f}_{n}}$ is a polynomial of degree n)

$f{{'}_{n}}\left( x \right)=n{{x}^{n-1}}-n=n\left( {{x}^{n-1}}-1 \right)$ but  $0\le x\le 1\Leftrightarrow 0\le {{x}^{n-1}}\le 1\Leftrightarrow -1\le {{x}^{n-1}}-1\le 0$

Hence $f{{'}_{n}}\left( x \right)<0$ $\Rightarrow {{f}_{n}}$ is strictly decreasing function thus monotone function

Also ${{f}_{n}}\left( 0 \right)=0-n0+1=1>0\,\,\And \,\,{{f}_{n}}\left( 1 \right)=1-n+1=2-n<0$

Hence ${{f}_{n}}\left( 0 \right)\times {{f}_{n}}\left( 1 \right)=1\left( 2-n \right)<0$ so by IVT ${{f}_{n}}\left( x \right)=0$ has unique solution ${{\alpha }_{n}}\in \left[ 0,1 \right]$

2) We have  ${{\alpha }_{n}}\in \left[ 0,1 \right]$ is a root for ${{f}_{n}}\Leftrightarrow {{f}_{n}}\left( {{\alpha }_{n}} \right)=0$ so we need to compute the following

${{f}_{n}}\left( \frac{1}{n} \right)={{\left( \frac{1}{n} \right)}^{n}}-n\frac{1}{n}+1=\frac{1}{{{n}^{n}}}>0$

And ${{f}_{n}}\left( \frac{2}{n} \right)=\frac{{{2}^{n}}}{{{n}^{n}}}-\frac{2n}{n}+1=\frac{{{2}^{n}}}{{{n}^{n}}}-1={{\left( \frac{2}{n} \right)}^{n}}-1$


We have $\frac{1}{n}<0\Leftrightarrow \frac{2}{n}<0\Leftrightarrow {{\left( \frac{2}{n} \right)}^{n}}<0\Leftrightarrow {{\left( \frac{2}{n} \right)}^{n}}-1<-1<0$

Thus $f\left( \frac{2}{n} \right)<{{f}_{n}}\left( {{\alpha }_{n}} \right)<f\left( \frac{1}{n} \right)\Leftrightarrow \frac{2}{n}<{{\alpha }_{n}}<\frac{1}{n}$

3) As $n\to \infty \,,\,\frac{2}{n}\to 0\,\,\And \,\,\frac{1}{n}\to 0$ hence by squeeze theorem ${{\alpha }_{n}}\to 0$

We have ${{\alpha }_{n}}$ is a root for ${{f}_{n}}$ thus ${{f}_{n}}\left( {{\alpha }_{n}} \right)=0\Leftrightarrow \alpha _{n}^{n}-n\alpha +1=0\Leftrightarrow n{{\alpha }_{n}}=\alpha _{n}^{n}+1$

Thus $\underset{n\to \infty }{\mathop{\lim }}\,n{{\alpha }_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( {{\alpha }_{n}} \right)}^{n}}+1=0+1=1$