Could someone explain something to me about the following question?
Find a $\delta >0$ such that if $|x|< \delta $, then $|\frac{1}{x+1}-1|<0.05$.
I got the inequality $-0.0476\leq x\leq 0.0526$, and the textbook chose $\delta = 0.0476$. Could someone please explain why they chose this $\delta$? Why not $\delta = -0.0476$? Would it be too big since then it would be like $0.04997<0.05$?
On
Note that to fulfill the condition obtained
$$-0.0476\leq x\leq 0.0526$$
we need to choose
$$\delta=\min\{0.0476,0.0526\}>0$$
such that for
$$|x|<\delta\implies -0.0476\leq x\leq 0.0526$$
indeed recall that
$$|x|<\delta\iff -\delta <x<\delta$$
On
You just have to find a $\delta$, not the best one. At least that is what you wrote.
$|\dfrac{1}{x+1}-1| =|\dfrac{1-(x+1)}{x+1}| =|\dfrac{x}{x+1}| $, so if $|x| < \frac12$, $|\dfrac{1}{x+1}-1| \lt \dfrac{|x|}{\frac12} =2|x| $.
Therefore, to make $|\dfrac{1}{x+1}-1| \lt .05$, it is sufficient to make $2|x| < .05$ or $|x| < .025$.
This is not the best, but it will do. If the problem wants the largest bound, it should say so.
Because that is a negative number.
And $|\frac 1{x+1} -1| < -0.0476 < 0$ will never happen.
And the didn't chose $\delta = 0.0526$ because that is too big/not strict enough.
The number $x = -0.05$ is such that $|x|< 0.0526$ but
$|\frac 1{x+1} - 1| = |\frac 1{-.05+1} -1| = |\frac 1{.95} - 1| = |1.0526315789473684210526315789474 - 1| = .0526315789473684210526315789474 > .05$.
If we have $-0.0476 < x < 0.0526$ then we have CAN'T have $x < -0.0476$, we must choose $\delta$ to be the MINIMUM of $|-0.0476|$ and $|0.0526|$.
Now you may be saying: but then we are losing all the $0.476 < x < 0.0526$ for which $x$ would be okay. But that's okay, we don't need find all possible $x$. Just a range of $x$ where $|\frac 1{x+1} - 1|$ must be less than $.05$. Not could be (as $0.476 < x < 0.0526$ could be) but must be (as $-0.0526< x < -0.0476$) can't be.