TITLE: Finding $\delta$ with given $\varepsilon$ for $f(x)=4x^2 +x +3$ and $x\to3$
QUESTION [7 upvotes]: $f(x)=4x^2 +x +3$ and the limit as x approaches $-3$ of $f(-3)= 36$, Find $\delta$ such that
$0<|x+3|<\delta \longrightarrow |f(x)-36|<.003$
I have tried:
$|(x+3)(4x-11)|<0.003$
$0<|x+3|<\frac{0.003}{|4x-11|}$
Assume $-4<x<-2$ 
$\delta= .000111$ or $0.000158$  Both came back as incorrect. Where have I gone wrong? 
EDIT: Problem solved via Henry's answer. Thank you all for the help! $\delta=0.000130$

REPLY [3 votes]: Let us see how you get the answer without using derivatives. What is the definition of continuity?
A function $f(x)$ is said to be continuous at $x=a$ if for all $\epsilon > 0$, there exists a $\delta >0$ such that 
$|x-a|< \delta$ implies that $|f(x) - f(a)| < \epsilon$.
So let $\epsilon > 0$ be given as in your problem ($\epsilon = 0.003$).
Now if $|x + 3| < \delta$, then we see that 
$-\delta < x+3 < \delta

\implies -4\delta - 23 < 4x - 11 < 4\delta - 23. $
The question is, "Now  that we have a bound on $4x - 11$ of the form $a < 4x - 11 < b$, how can I make it into the form $-c < 4x - 11 <c$?" $a,b$ and $c$ are just some real numbers.
The reason I would like to this is in order for me to put a bound on the absolute value of $4x - 11$.
Notice that $4\delta - 23 < 4\delta + 23$, so that 
$-(4\delta + 23) < 4x - 11 < 4\delta - 23 < 4\delta + 23$.
In other words you can be confident now that
$|4x - 11| < 4\delta + 23$.
So if you consider $|f(x) - 36|$, you see that  $|x +3| < \delta$ will lead to the conclusion that
$|f(x) - 36|= |(x+3)(4x - 11)| < \delta(4\delta + 23)$.
If you solve $\delta(4\delta + 23) = \epsilon = 0.003$ for delta, you will get
$\delta = 0.0001304$ or $\delta = -5.7501304$. Since delta is positive, you will only have one choice of delta, which is 0.0001304 and you're done!
No guessing, playing around with numbers, using derivatives and working blindly.