Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta
I am having difficulty finding the $\delta$ value.
Here is what I have done so far:
What I want to show:
$$\forall \epsilon > 0, \exists \delta > 0 \; such \; that \; 0<\mid x -2 \mid < \delta \implies \mid x^4 - 2x^3 + x - 2\mid < \epsilon $$
After a lot of manipulation, I was able to turn the equation to the following:
$$\mid x^4 - 2x^3 + x - 2\mid \; = \; \mid x-2 \mid \mid x-2 \mid \mid x + 1 \mid \mid x + 1 \mid + 3\mid x + 1 \mid \mid x - 2 \mid$$
I already know that $ \mid x -2 \mid < \delta$. I must find an upper bound for $\mid x + 1 \mid$.
$$\mid x + 1 \mid = \mid x + 1 -3 + 3 \mid \; \le \; \mid x - 2 \mid + 3 $$
Choose $\delta = 1$, then
$$\mid x - 2 \mid < 1 \implies \mid x -2 \mid + 3 < 4$$
Therefore $\mid x + 1 \mid < 4$
This means:
$$\mid x^4 - 2x^3 + x - 2\mid < (\delta)(\delta)(4)(4) + 3(\delta)(4) = 16\delta^2 + 12\delta = \epsilon$$
This is where I get stuck. How do I transform $16\delta^2 + 12\delta = \epsilon$ into solving for $\delta$ in terms of $\epsilon$?
I am not sure if I am over thinking the whole question or if the above is even remotely correct. Please let me know any mistakes that I have made.
 A: Since you have done all this work so far, just use the quadratic formula to complete the argument. The two roots are $\frac{ -12 \pm \sqrt{(-12)^2 + 4(16)(\epsilon)}}{2(16)}$. Then set 
$$\delta = \min\left\{1,\frac{-12 +\sqrt{144 +64\epsilon}}{32}\right\}$$
The equation always has real roots. I had to take the minimum of $1$ and the root since you had already made the assumption $\delta<1$ in your computations.
A: It can be settled without resorting to the "quadratic formula". 
$16\delta^2+12\delta = 4\delta(4\delta+3)$. If you choose $\delta < 1$, then your job is a bit easier, since then $4\delta+3 < 7$, and $4\delta(4\delta+3) < 7\cdot 4\delta = 28\delta$. Next you want that $28\delta < \epsilon $, thus $\delta < \dfrac{\epsilon}{28}$. Now you have an idea of how to select $\delta$. You can take $\delta = \text{min}(1,\frac{\epsilon}{28})$.
A: Here's a very easy and brief way to do it:
Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-2|<\delta$, then $|(x^4-2x^3+x+3)-5|<\epsilon$. 
Now, 
$$|(x^4-2x^2+x+3)-5|=|(x-2)(x^3+1)|.
$$ If $|x-2|<1$, that is, $1<x<3$, then $x^3+1<3^3+1=28$, and so 
$$
|(x^4-2x^3+x+3)-5|=|x-2|(x^3+1)<28|x-2|.
$$ So if we take $\delta=\min\left\{1,\frac{\epsilon}{28}\right\}$, then 
$$
0<|x-2|<\delta\to|(x^4-2x^2+x+3)-5|=|(x-2)|(x^3+1)<\frac{\epsilon}{28}\cdot 28=\epsilon.
$$ Thus, by the definition of a limit, $\lim_{x\to 2}(x^4-2x^3+x+3)=5$.
A: 
$$\text{Write the polynomial as a sum of powers of $x-2$.}$$

$$x^4-2x^3+x+3=(x-2)^4+6(x-2)^3+12(x-2)^2+9(x-2)+5$$

The most efficient way to do this is to apply Ruffini-Horner repeatedly with the polynomial with the value $x=2$. 

Then $$|x^4-2x^3+x+3-5|=|(x-2)^4+6(x-2)^3+12(x-2)^2+9(x-2)|\leq(1+6+12+9)|x-2|$$
when $|x-2|<1$. So, you only need to put $\delta=\min(1,\epsilon/(1+6+12+9)).$
