Picking a small value for $\delta$ to prove function limit by $\epsilon$-$\delta$ definition I'm having difficulty understanding the common textbook way of starting from a random small value(often 1) for $\delta$. For example, to prove $f(x)=x^2+x-3 \rightarrow -1 \;as\; x \rightarrow 1$, textbooks often start by setting $\delta <= 1$, then plug in this inequality into the function to get $4\delta < \epsilon$. Isn't doing so only proving for only the $\epsilon$ whose corresponding $\delta$ is less than or equal to 1 intead of for every $\epsilon$ and every corresponding $\delta$? 
 A: Your are not prroving anything for every $\delta$. You are required to prove that for every $\epsilon >0$ there exists $\delta >0$ such that something happens. $\delta$ is completely at your choice and it is very often convenient to take $\delta <1$. 
A: I think the real problem here, faced by many students, is the fact that the proof of the result is written in a specific order BUT we find the information for the proof (eg in this case what we need δ to be) in a different order.  
To find the relevant δ we use the fact that our final line needs to be less than ε and work backwards.  I teach my students to use the standard proof format but to leave blanks that we return to and complete once we know the information!
In more complicated situations such as this one, look for the |x-a| term and keep it for the |x-a|< δ and then look for a constant value that will give you a bound (any bound) on the other terms.  You then choose the minimum of all the values used.  
Remember our result has to work for any ε, the "hard" εs are the small ones so we often think of ε as being small, but our result has to work when ε is large as well!
A: Your proof should start with: "let $\epsilon>0$". Then let $x$ such that $|x-1|<\delta$ for some $\delta>0$ (to be determined as a function of $\epsilon$) and consider the difference
$$|f(x)-f(1)|=|x^2+x-2|=|x+2|\cdot |x-1|\leq |x+2|\delta.$$
Now, try to find $\delta>0$ such that
$$|x+2|\delta<\epsilon$$
for all $x$ such that $|x-1|<\delta$.
