Show that $|f(p_n)|1$ but that $|p-p_n|1000$. Let $f(x)=(x-1)^{10}$.

The root of the equation , $p=1$.

The approximates of the root, $p_n=1+\frac{1}{n}$

Show that $|f(p_n)|&lt;10^{-3}$ whenever $n&gt;1$ but that $|p-p_n|&lt;10^{-3}$ requires that $n&gt;1000$.

I started to solve it by the following way:

$|p-p_n|&lt;10^{-3}\Rightarrow |1-(1+\frac{1}{n})|&lt;10^{-3}\Rightarrow |\frac{1}{n}|&lt;10^{-3}\Rightarrow |n|&gt;1000\Rightarrow n&gt;1000$

And

$f(p_n)=[(1+\frac{1}{n})-1]=\frac{1}{n}$; whenever $n&gt;1$

But how to show 

$|f(p_n)|&lt;10^{-3}$ ?
You forgot the power of $10$ when you computed $f(p_n)$.
Show that $|f(pn)|&lt;10^{-3}$ whenever $n&gt;1$
$F(x)=(x-1)^{10}$ then $f(pn) =((1+1/n)-1)^{10} =&gt; |f(pn)| =(1/n)^{10}$
And $n&gt;1$ then $|f(pn)| &lt; = (1/2)^{10} = 1/1024$ and $1/1024 &lt;10^{-3 }$