Calculus - Finding suitable delta I want to understand the concept of choosing $\delta$ since it doesn't make sense to me, I hope someone could simplify it.
I'll show an example to the problem
I have this polynom :
$$f(x)=4x^3+3x^2-24x+22$$
I want to show that this polynom is continuous at x=1, since f(1)=5 :
$$\lim_{x \to 1} (4x^3+3x^2-24x+22)=5$$ 
more specifically,
To all $\epsilon$>0 exist $\delta$>0 so to $x$∈R (function is a polynomial), $$|x−1|<\delta \mbox{ then } |4x^3+3x^2-24x+22-5|<\epsilon$$
Ok, so I simplify the expression :
$$|4x^3+3x^2-24x+22-5|=$$
$$|4x^3+3x^2-24x+17|=$$
$$|(x-1)(4x^2+7x-17)|=$$
$$|x-1|*|4x^2+7x-17|=$$
Ok, now I need to choose $\delta$, lets assume I choose a huge $\delta$ 
which is not suitable it a little bit hard to explain it, Lets assume that the case in the picture happens:
http://oi57.tinypic.com/2642m10.jpg
(I'm aware that the function is not well painted, did my best)
Lets assume that $\delta$=1000 is not suitable as the case shown in the picture and solve this problem.. :
$$|x−1|<1000$$
$$-1000<x-1<1000$$
$$-999<x<1001$$
Therefore :
$$|x-1|*|4x^2+7x-17|\leq|x-1|*|4*1001^2+7*1000-17|=|x-1|*4014987$$
Then: $\delta$=min{$1000,\frac{\epsilon}{4014987}$}
First of all, I wonder if my answer is correct since I took unsuitable $\delta$, If my answer is correct how is it possible eventhough I took unsuitable $\delta$?
This question is in case that my answer is not correct :
My point is when choosing a small like: $\delta=1$ it will probably be "suitable" since it little, but there might be a function that small $\delta$ like $\delta=1$ will be "unsuitable" how could I ensure that the $\delta$ I choose it is suitable for specific function?
I'm little confused by that issue, I hope somone could explain also intuitive explation will be great.
Thanks.
 A: In the end the choice was $\delta=min\{1000,\epsilon/4014987\}$ which is very small for a small $\epsilon$ like 1.
The first 'choice' of   δ ,is "unsuitable" for example, in $$\lim_{x \to 1} (1/x)=1$$ ,for δ>1,because then no ϵ will fit.
This choice is made for 2 reasons: 


*

*to narrow the slop of function we are talking about,to avoid impossible (as $$\lim_{x \to 1} (1/x)=1$$) or difficult parts.Here the main concept is that it does not matter if we take a 'too small' δ. We do this in functions with un-continous points.

*To make it easy to evaluate the function,as you did here.In this case,a big δ does not make any harm as long as we can still evaluate the function.
A good δ,in a case of a polynom,is one that makes the function easy to evaluate,witch means,easy to find the sup and inf,or bound them.so it would be nice to take a δ that gives us  a parabulaa looking slop.
Note that the "To all ϵ>0 exist δ>0 so..." from the limit condition,says that the δ depends on the specific ϵ , so in the end of the proof we always need to make a min{} choice(exept constent functions, where δ can remane const.)
A little mistake is $|4x2+7x−17|≤|x−1|∗|4∗10012+7∗1000−17|$ which is true in this case,but you cant always just take the biggest option for x (somtimes a smaller negative x,gives a larger value,for example).If we are dealing with a polynom,we just need to find thr maximum falue (in this case  in $ -999<x<1001$)
A: $$ |f(x)-f(1)|<\varepsilon \Leftrightarrow | (x-1)(4x^2+7x-17)|<\varepsilon.$$
Let $|x-1|<\delta$ so: $$  | (x-1)(4x^2+7x-17)|<\delta |4x^2+7x-17|$$.
Pick $\delta\le 1$ so $|x-1|<\delta$ implies $|x|<2$ which in turn implies:
$$\delta |4x^2+7x-17|< \delta (|4x^2+7x|+17)<\delta (16+14+17)=37\delta.$$
So let $\delta=\min\{1,\frac\varepsilon{38}\}$.
Now $|x-1|<\delta=\min\{1,\frac\varepsilon{38}\}$ implies $|f(x)-f(1)|<37\cdot\delta<\varepsilon$.
