2
$\begingroup$

My question concerns the following proof:

Prove that $$\lim_{x\to -5} \vert x - 5 \vert=10.$$

We are given $\varepsilon > 0$. Then we have $$\begin{align*} \vert \vert x - 5 \vert - 10 \vert &< \varepsilon \\ \vert -(x-5)-10 \vert &< \varepsilon \qquad \text{($x-5 < 0$)} \\ \vert (-x-5) \vert &< \varepsilon \\ \vert (x-(-5)) \vert &< \varepsilon \end{align*}$$

So, let $\delta = \varepsilon$.

So for $\vert x - (-5) \vert < \delta = \varepsilon$, you have

$$\begin{align*} \vert-(x+5) \vert &< \varepsilon \\ \vert -(x-5)-10 \vert &< \varepsilon \\ \vert \vert x - 5 \vert - 10 \vert &< \varepsilon \qquad \text{(because $x-5 < 0$)}. \end{align*}$$

Now surely I'm missing something very obvious here, but why are we allowed to make the claim that $x-5 < 0$, or is that a restriction we are imposing? It seems if we wanted to make this claim we would need to include in the proof something like: assume $x < 5$?

$\endgroup$
  • $\begingroup$ We are not allowed to assume that x-5<0. In fact you can use your proof with the case x-5>0. It works similarly. $\endgroup$ – mapping Feb 27 '17 at 16:06
  • $\begingroup$ Did you write the attempted proof that you asked us to check, or are you asking us to explain something that someone else wrote? If it's the latter (which would have been my guess, since you didn't seem to know why one would write $x-5<0$), it's not actually "my attempt" (but the question is fine, aside from that). $\endgroup$ – David K Feb 27 '17 at 17:02
4
$\begingroup$

You're right, that is a restriction we are imposing. We want $$|x-(-5)| < \delta \Rightarrow ||x-5| - 10| < \varepsilon.$$ Note that if you find some $\delta$ that works, than any $\delta'\leq\delta$ will also do. What we could do is take $\delta = \min\{\varepsilon, 10\}$. Then,

  • $|x + 5| < \delta \leq 10$, so $x-5<0$, and
  • $|x + 5| < \delta \leq \varepsilon$,

which are the two restrictions on $x$ you needed. Now the rest of your argument follows.

$\endgroup$
1
$\begingroup$

Let $|x+5|<1$ and for given $\varepsilon > 0$, there is $\delta>0$ such that if $|x+5|<\delta$ then $$||x - 5|- 10 = |\sqrt{(x-5)^2}-10|=\dfrac{|x+5|~|x-15|}{\sqrt{(x-5)^2}+10}< \dfrac{21|x+5|}{10}<\dfrac{21}{10}\delta$$ take $\delta = \min\{1,\dfrac{10}{21}\varepsilon \}$.

$\endgroup$

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.