9
$\begingroup$

For the limit $$\lim_{x\to 5}\sqrt{x-1}=2$$ find a $\delta>0$ that works for $\epsilon=1$.

In another words, find a $\delta>0$ such that for all $x$, $$0<|x-5|<\delta \implies |\sqrt{x-1}-2|<1$$


Ok so here's what I did... $$|\sqrt{x-1}-2|<1$$ $$-1<\sqrt{x-1}-2<1$$ $$1<\sqrt{x-1}<3$$ $$1<x-1<9$$ $$2<x<10$$
Since I have to get $x$ in the simplified inequality above to change to $x-5$, I decided to subtract 5 on all sides like this: $$2-5<x-5<10-5$$ $$-3<x-5<5$$

But I don't know if that's right or not. Please help?

$\endgroup$
  • $\begingroup$ Try it out! Plug in different values for your $\delta$ from the range you found. $\endgroup$ – The Count Jun 27 '16 at 19:48
7
$\begingroup$

Assuming your algebra is correct (I didn't check it carefully), you have shown that $|\sqrt{x-1}-2|<1$ if and only if $-3<x-5<5$. Now you want to find $\delta>0$ so that if $|x-5|<\delta$ then $-3<x-5<5$. But $|x-5|<\delta$ is the same as $-\delta<x-5<\delta$. So what is a suitable value of $\delta$?

$\endgroup$
  • $\begingroup$ Yes, that's right $\endgroup$ – Aniket Jun 27 '16 at 21:13
8
$\begingroup$

You know that if $-3 < x-5 < 3 \implies -3 < x-5 < 5$. This means you can take $\delta = 3$.

$\endgroup$
  • $\begingroup$ Yea the answer is 3, but how did you directly get -3 < x - 5 < 3 from -3 < x - 5 < 5? $\endgroup$ – Aniket Jun 27 '16 at 21:11
  • $\begingroup$ It comes from $-3$, then I think the other is $3$. $\endgroup$ – DeepSea Jun 27 '16 at 21:14
  • $\begingroup$ Can't we take $-5$ and $5$, instead of $-3$ & $3$? $\endgroup$ – Mr Reality Jul 31 '18 at 2:08

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.