Guessing delta value, in epsilon-delta proof So as I've solved various problems regarding epsilon-delta proof, I have faced several questions where I had to kind of implement this process:
So here's one of them:
$\lim_{x\rightarrow 1} (x^3+x+1) = 3 $
So I started off saying,
Let $\epsilon$ > 0 be given. We want to find $\delta$ > 0 s.t. if 0 < $|x-1|$ < $\delta$, then $|x^3+x+2| < \epsilon$. 
$|x^3+x+2| = |(x-1)(x-1)(x+2)|$  ..... (a)
(Here comes the part! ) 
Let $\delta < 1, $ ... (b)
$ |x-1| < 1$ 
$-1 < x-1 < 1$
$1 < x + 2 < 3$
$x + 2 < 3$
Apply to the part (a), $<(x-1)\times 1 \times 3  < \delta \times 3
 = \epsilon$ 
$\delta = \epsilon/3$
$\delta = min(1, \epsilon/3)$.
and the proof follows...
So my question is what is the standard of choosing n in part (b) Let $\delta < n$? 
Sometimes they choose $\frac{1}{2}$ or $\frac{1}{4}$ for n, and sometimes 1.
For this case, my answer sheet specifically chose $\frac{1}{2}$ for n and I have no idea why.
 A: The key idea is  to pick it to be sufficiently small, avoiding technical singularity, aim for simplicity to achieve your goal.
$1$ is a popular choice but it need not work all the time. We need to pick it to be small enough to have control over the whole function. 
For example, if you want to prove that $$\lim_{x \to 0.5}\frac1x=2$$ 
For any $\epsilon >0$, we want to show that we can find $\delta >0$, such that if $0<|x-0.5|<\delta$, then $|\frac1x-2|<\epsilon$.
To get ideas on how to pick $\delta$, I would  work backward on a draft paper and write stuff like $$\frac{|1-2x|}{|x|}<\epsilon$$
If $|x-0.5|<\delta$, then $|1-2x|<2\delta$, that is we know how to bound the numerator. 
For denominator, we want to make things simpler and we try to bound $\frac1x$ by a positive number. If we insist that $|x-0.5|<1$, then it is possible that $x=0$, and we know there is a singularity. Hence, avoid it by picking a smaller number, say pick $|x-0.5|<0.25$, then $0.25<x < 0.75$, then $\frac1{|x|}<4$
Hence, we can pick $\delta = \min\left(\frac14,  \frac{\epsilon}8 \right)$, then
$$\frac{|1-2x|}{|x|}<8\delta \le \epsilon.$$
A: Before really answering the question, I remark that OP has made a careless mistake in (a) while recalling the epsilon-delta definition of limits.

Let $\epsilon$ > 0 be given. We want to find $\delta$ > 0 s.t. if 0 < $|x-1|$ < $\delta$, then $|x^3+x+2| < \epsilon$.

Here it should be $|x^3+x+1-3| = |x^3+x-2|$ instead of $|x^3+x+2|$.
A quick short division tells us that $x^3+x-2 = (x-1)(x^2+x+2)$.
At the end of the epsilon-delta definition for the particular limit, we see "$|x^3 + x - 2|<\epsilon$".  Using $0<|x-1|<\delta$, where $\delta>0$ is to be determined, we get $|x^2+x+2|\,\delta < \epsilon$.  From this, it's clear that what we need is a positive upper bound $L$ for $|x^2+x+2|$, so that $\delta < \epsilon / L$ would imply $|x^3 + x - 2|<\epsilon$.  Here, $L$ can depend on $\epsilon$, but not $\delta$, since in the epsilon-delta definition, the existence of $\delta$ comes after that of $\epsilon$.
To explore the boundedness of $x^2 + x + 2$ for $x \in (1-\delta,1+\delta)$, without considering the properties of a quadratic polynomial, we might want to argue that
$$(1-\delta)^2 < x^2 < (1+\delta)^2 \tag1 \label1$$
from
$$1 - \delta < x < 1 + \delta. \tag2$$
But this requires $1 - \delta \ge 0$, i.e. $\delta \le 1$.  After that, you might want to add them up to get
$$\delta^2 - 3\delta + 4 < x^2+x+2 < \delta^2 + 3\delta + 4.$$
As a result, $$|x^2+x+2| \le \max\{\delta^2 - 3\delta + 4,\delta^2 + 3\delta + 4\} = \delta^2 + 3\delta + 4$$ for any $0 < |x-1| < \delta$.  Now the above upper bound for $|x^2+x+2|$ depends on $\delta$, which violates the last sentence of the previous paragraph.  To logically validate the argument, just replace all instances of $\delta$ with another constant $d$ in this paragraph, so that $L$ is solely a function of $d$ (and that $L$ is independent from $\delta$).
Now, let's get to the heart of your question.

What is the standard of choosing $n$ in part (b) "Let $\delta<n$"?
  Sometimes they choose $1/2$
  or $1/4$ for $n$, and sometimes $1$. For this case, my answer sheet specifically chose $1/2$ for $n$ and I have no idea why.

Before continuing the discussion, I remark that $n$ often connotes natural numbers, so it's more sensible to rename it.  In the following, I'll rename $n$ as $d$, which corresponds to the variable $d$ in the above paragraph.  To understand why a particular value of $d$ is chosen, it's important to know how a chosen value $d$ affects the value of the upper bound $L$.  Now, recall that $0 < |x-1| < \delta < d$, so the arguments in the above paragraph can be applied to (b) to conclude that
$$\forall x \in (1-d,1+d), |x^2+x+2| < d^2 + 3d + 4 =: L(d).$$
Therefore, for any $d \in (0,1]$, put $L(d) = d^2 + 3d + 4$ and choose $\delta = \min \{d, \epsilon / L(d)\}$, so that
$$|x^3 + x - 2| = |x-1|\,|x^2 + x + 2| < \delta L(d) = (\epsilon / L(d)) \, L(d) = \epsilon.$$  Here $d > 0$ since it represents the radius of a neighbourhood, and $d \le 1$ is a technical requirement so that $x$ stays positive for \eqref{1} to be valid.
Observe that $L(d) = d^2 + 3d + 4 = (d + 3/2)^2 + 7/4$, so it's a parabola opening upwards with vertex $(-3/2, 7/4)$.  This show that $L(d)$ strictly increases on $d \in (0,1]$.  For any fixed $\epsilon > 0$, as $d$ increments from $0$ to $1$, $L(d)$ also increases, so $\delta$ would decreases if the error $\epsilon$ is small enough.
Conclusion: for the epsilon-delta proof of this limit to work, you may choose any $d \in (0,1]$.  Choosing $d = 1$ will simplify the calculations, while a smaller value $d$ will lower the value of $L(d)$, so that it's easier to hit $x \in (1-\delta, 1+\delta)$ for any fixed sufficiently small error $\epsilon > 0$.
Remarks: Intuitively, due to the Extreme Value Theorem and the Intermediate Value Theorem, the continuous function $x \mapsto x^2 + x + 2$ maps a closed and bounded interval to another closed and bounded interval.  Since you won't logically change the epsilon-delta definition by replacing $|x-1| < \delta$ with $|x-1| \le \delta$, $x^2 + x + 2$ is bounded on for $x \in [1-\delta,1+\delta]$.  To solve the dependency problem of $L$, just replace $\delta$ with any positive constant $d$ you like, so that $0<|x-1|\le d$ gives $|x^2 + x + 2| \le L(d)$.  As the definition of continuity depends on limits, which is defined by epsilon-delta, I leave this paragraph in the remarks.
