I am working through the Stewart Calculus text independently and am stuck on one of the practice problems (edition 7e - problem 29 in section 1.7).
I am confused by particularly by hint #3 which says:
Why is $\lvert x^2 -4x + 4 \rvert = \lvert x + 2 \rvert \lvert x - 2 \rvert$
I am quite confused by this and I cannot figure out how these two sides of the equation are equal. For full context of the homework hints please see here (I would have attached the image but I cannot since I have not posted enough).
Left to my own device I would have figured that $\lvert x^2 -4x + 4\rvert \lt \epsilon $ where the left hand sides could be reworked as $ \lvert (x -2)^2 \rvert \lt \epsilon $ and furthermore that $ x - 2 = \sqrt{\epsilon}$
Not sure where I am going wrong here, can somebody please help?
Solution
Suppose that $|x - 2| \leq \delta$. Then one has that \begin{align*} |f(x) - f(2)| & = |x^{2} - 4x + 5 - 1|\\\\ & = |x^{2} - 4x + 4|\\\\ & = |x-2|^{2} \leq \delta^{2} = \varepsilon \end{align*}
Thus we can conclude that for every $\varepsilon > 0$, there corresponds $\delta = \sqrt{\varepsilon}$ such that \begin{align*} |x - 2| \leq \delta \Rightarrow |f(x) - f(2)| \leq \varepsilon \end{align*}
and we are done.
If you still have any doubts, please let me know.
Comments
The proposed identity is not correct. Indeed, $|x^{2} - 4x + 4| \neq |x-2||x+2|$.