TITLE: Finding Delta Algebraically for a Given Epsilon?
QUESTION [10 upvotes]: 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?

REPLY [9 votes]: You know that if $-3 < x-5 < 3 \implies -3 < x-5 < 5$. This means you can take $\delta = 3$.