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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

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