# Finding Delta Algebraically for a Given Epsilon?

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: $$2-5<x-5<10-5$$ $$-3<x-5<5$$

• Try it out! Plug in different values for your $\delta$ from the range you found. Jun 27, 2016 at 19:48

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 $\delta$?
You know that if $-3 < x-5 < 3 \implies -3 < x-5 < 5$. This means you can take $\delta = 3$.
• It comes from $-3$, then I think the other is $3$. Jun 27, 2016 at 21:14
• Can't we take $-5$ and $5$, instead of $-3$ & $3$? Jul 31, 2018 at 2:08
• @MrReality No because we want a tight bound to imply a looser bound. i.e. $-3<x-5<3$ implies $-3<x-5<5$. But a loose bound doesn't imply a tight bound. i.e. $-5<x-5<5$ doesn't imply $-3<x-5<5$ (take $x=1$ for example).