3
$\begingroup$

I have an assignment about epsilon-delta proofs and I'm having trouble with this one. I have worked it through using some of the methods I've picked up for similar proofs but it's just something about this particular expression that doesn't sit right with me. Any feedback would be very helpful. This is how far I've come:

Let $\varepsilon > 0$. We want to find a $\delta$ so that $\left|\frac{1}{x - 1} - 1\right| < \varepsilon$ when $0 < |x - 2| < \delta$. We expand the expression: \begin{align*} \left|\frac{1}{x - 1} - 1\right| &< \varepsilon \\ \left|\frac{1}{x - 1} - \frac{x - 1}{x - 1}\right| &< \varepsilon \\ \left|\frac{2 - x}{x - 1}\right| &< \varepsilon \\ |{x - 1}| &< \frac{|x - 2|}{\varepsilon} \\ \end{align*}

We could let $\delta = \dfrac{|x - 2|}{\varepsilon}$ but $|x - 2|$ contains an unwanted variable. Since the limit is only relevant when $x$ is close to $a$ we'll restrict $x$ so that it's at most $1$ from $a$ or in other words, in our case, that $|x - 1| < 1$. This means $0 < x < 2$ and that $-2 < x - 2 < 0$. Looking at our previous inequality

\begin{align*} |{x - 1}| &< \frac{|x - 2|}{\varepsilon} \end{align*}

we see that the right-hand side is the smallest when $|x - 2|$ is the smallest which by the range above is when $x - 2 = -2$ and then we have that

\begin{align*} |{x - 1}| &< \frac{|x - 2|}{\varepsilon} < \frac{2}{\varepsilon} \end{align*}

We now have the two inequalities $|x - 1| < 1$ and $|x - 1| < \frac{2}{\varepsilon}$. Let $\delta = \textrm{min}(1, \frac{2}{\varepsilon})$ and by definition we have that for every $\varepsilon > 0$ there is a $\delta$ so that $|f(x) - A| < \varepsilon$ for every $x$ in the domain that satisfies $0 < |x - a| < \delta$. $\blacksquare$

$\endgroup$
  • $\begingroup$ How did you get to that last inequality for $x-1$. That doesn't seem right since $x-1$ is in the denominator $\endgroup$ – imranfat Oct 2 '19 at 23:03
  • $\begingroup$ the last inequality should be $$|x-1|>\frac{|x-2|}\varepsilon.$$ $\endgroup$ – YiFan Oct 2 '19 at 23:06
  • $\begingroup$ Is it because the reciprocal reverses the inequality? I wasn't too sure about that operation it seems… $\endgroup$ – Markus Amalthea Magnuson Oct 2 '19 at 23:19
3
$\begingroup$

Assuming wlog $\frac32\le x\le \frac52$ we have

$$\left|\frac{2 - x}{x - 1}\right| < \varepsilon \iff \left|2 - x\right| < \varepsilon \left|x - 1\right|\le \frac 32 \varepsilon $$

then it suffices to take $\delta <\frac 32 \varepsilon $.

$\endgroup$
  • $\begingroup$ Thank you, this is very straightforward. For some reason, we rarely talk about "without loss of generality" in our lectures etc. so I was wondering in this case if we can say wlog because values for $x$ that are close to $a$ still give us the same result when talking about limits since they approach $a$? $\endgroup$ – Markus Amalthea Magnuson Oct 2 '19 at 23:22
  • $\begingroup$ Thanks I fix the typo! $\endgroup$ – user Oct 3 '19 at 3:52
  • $\begingroup$ Yes we can say wlog because we are interested to values for 𝑥 that are close to a, and the restriction still give us the same result. $\endgroup$ – user Oct 3 '19 at 3:56
  • $\begingroup$ @JohnOmielan I lost points on this with the comment "You need to approximate 1/|x-1|". Why do you think that is? I'm trying to figure this out before asking the assistant who graded it. $\endgroup$ – Markus Amalthea Magnuson Oct 21 '19 at 11:12
  • $\begingroup$ @MarkusAmaltheaMagnuson I don't know exactly what you submitted, so it's hard for me to guess why the grading assistant may have wrote that comment. If you used what you wrote in your question, along with finishing it such as suggested by the solution above, your proof should have been complete & correct. Did you perhaps skip anything? I'm sorry I can't be of more help. $\endgroup$ – John Omielan Oct 21 '19 at 16:47
2
$\begingroup$

Hint: You may suppose, w.l.o.g. that $|x-2|<\frac12$, in which case $|x-1|>\frac12$, so that $$\left|\frac{x-2}{x - 1}\right| <2|x-2|.$$

$\endgroup$
0
$\begingroup$

$\delta<\min(1,\varepsilon)/2$ should do the job.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.