Pointwise Convergence .... Abbott, Example 6.2.2 (iii) .... ....

[Math Amateur]

I am reading Stephen Abbott's book: "Understanding Analysis" (Second Edition) ...

I am focused on Chapter 6: Sequences and Series of Functions ... and in particular on pointwise convergence...

I need some help to understand the 'mechanics' of Example 6.2.2 (iii) ...

Example 6.2.2 reads as follows:View attachment 8580In the above text from Abbott, in Example 6.2.2 (iii), we read the following:


" ... ... $\displaystyle \lim_{n \to \infty }$ \(\displaystyle h_n (x) = x\) $\displaystyle \lim_{n \to \infty }$ \(\displaystyle x^{ \frac{1}{ 2n -1 } } = \ \mid x \mid \) ... ... "

Can someone please explain exactly how ...\(\displaystyle x \) $\displaystyle \lim_{n \to \infty }$ \(\displaystyle x^{ \frac{1}{ 2n -1 } } = \ \mid x \mid \) ... ... "
Peter

 

[Chris L T521]

I am reading Stephen Abbott's book: "Understanding Analysis" (Second Edition) ...

I am focused on Chapter 6: Sequences and Series of Functions ... and in particular on pointwise convergence...

I need some help to understand the 'mechanics' of Example 6.2.2 (iii) ...

Example 6.2.2 reads as follows:In the above text from Abbott, in Example 6.2.2 (iii), we read the following:


" ... ... $\displaystyle \lim_{n \to \infty }$ \(\displaystyle h_n (x) = x\) $\displaystyle \lim_{n \to \infty }$ \(\displaystyle x^{ \frac{1}{ 2n -1 } } = \ \mid x \mid \) ... ... "

Can someone please explain exactly how ...\(\displaystyle x \) $\displaystyle \lim_{n \to \infty }$ \(\displaystyle x^{ \frac{1}{ 2n -1 } } = \ \mid x \mid \) ... ... "
Peter

Keep in mind that $x\in[-1,1]$. Note that
\[\lim_{n\to\infty} x^{\frac{1}{2n-1}}=\begin{cases}-1, & x\in[-1,0),\\ 0, & x=0,\\ 1, & x\in(0,1].\end{cases}\]
So in the end, we find that
\[x\lim_{n\to\infty}x^{\frac{1}{2n-1}} = \begin{cases}x, & x\geq 0,\\ -x,& x<0\end{cases} = |x|.\]
I hope this makes sense!

 

[Math Amateur]

Keep in mind that $x\in[-1,1]$. Note that
\[\lim_{n\to\infty} x^{\frac{1}{2n-1}}=\begin{cases}-1, & x\in[-1,0),\\ 0, & x=0,\\ 1, & x\in(0,1].\end{cases}\]
So in the end, we find that
\[x\lim_{n\to\infty}x^{\frac{1}{2n-1}} = \begin{cases}x, & x\geq 0,\\ -x,& x<0\end{cases} = |x|.\]
I hope this makes sense!

Thanks Chris ...

... just a clarification ...

How did you determine that

\(\displaystyle \lim_{n\to\infty} x^{\frac{1}{2n-1}} = -1, \ \ \text{ for } \ \ x\in [-1,0)\)

and

\(\displaystyle \lim_{n\to\infty} x^{\frac{1}{2n-1}} = 1, \ \ \text{ for } \ \ x \in (0,1]\) ...Do you just have to know these limits ... ?PeterPS I'm particularly curious about the case where x is negative ...

 

[Chris L T521]

Thanks Chris ...

... just a clarification ...

How did you determine that

\(\displaystyle \lim_{n\to\infty} x^{\frac{1}{2n-1}} = -1, \ \ \text{ for } \ \ x\in [-1,0)\)

and

\(\displaystyle \lim_{n\to\infty} x^{\frac{1}{2n-1}} = 1, \ \ \text{ for } \ \ x \in (0,1]\) ...Do you just have to know these limits ... ?PeterPS I'm particularly curious about the case where x is negative ...

The power on $x$ is $\frac{1}{2n-1}$ which implies these are odd radicals for $n\geq 2$ (when $n=1$, the expression is just $x$). So if $x>0$, $x^{\frac{1}{2n-1}}>0$ and since $\frac{1}{2n-1}\rightarrow 0$ as $n\rightarrow\infty$, we find that $x^{\frac{1}{2n-1}}\rightarrow 1$ (since $a^0=1$ for $a\neq 0$).

On the other hand, when $x<0$, we can factor negatives out of odd radicals, so we find that $x^{\frac{1}{2n-1}}=-(-x)^{\frac{1}{2n-1}}<0$. Since $x<0\implies -x>0$, we find that as $n\to\infty$, $x^{\frac{1}{2n-1}}=-(-x)^{\frac{1}{2n-1}}\rightarrow -1$ for $x<0$ (note that $(-x)^{\frac{1}{2n-1}}\rightarrow 1$ as $n\rightarrow \infty$ by similar reasoning in the previous paragraph).

I hope this clarifies things!

 

[Math Amateur]

The power on $x$ is $\frac{1}{2n-1}$ which implies these are odd radicals for $n\geq 2$ (when $n=1$, the expression is just $x$). So if $x>0$, $x^{\frac{1}{2n-1}}>0$ and since $\frac{1}{2n-1}\rightarrow 0$ as $n\rightarrow\infty$, we find that $x^{\frac{1}{2n-1}}\rightarrow 1$ (since $a^0=1$ for $a\neq 0$).

On the other hand, when $x<0$, we can factor negatives out of odd radicals, so we find that $x^{\frac{1}{2n-1}}=-(-x)^{\frac{1}{2n-1}}<0$. Since $x<0\implies -x>0$, we find that as $n\to\infty$, $x^{\frac{1}{2n-1}}=-(-x)^{\frac{1}{2n-1}}\rightarrow -1$ for $x<0$ (note that $(-x)^{\frac{1}{2n-1}}\rightarrow 1$ as $n\rightarrow \infty$ by similar reasoning in the previous paragraph).

I hope this clarifies things!

Very clear thanks Chris ...

Your help is much appreciated ...

Peter