Finding a "suitable" $\delta$ given a limit Find a suitable $\delta$ for the following limit.
$$\lim_{x\to -2} (2x^2 + 9x - 1) = -11$$
I thought that by "suitable" it just means that I pick a random $\epsilon$ and then find the $\delta$ corresponding to it?
So I go with $\epsilon = .01$
$-11-.01 < f(x) < -11+.01$
$-11.01 < 2x^2 + 9x - 1 < -10.99$
$-10.01 < 2x^2 + 9x < -9.99$
after some rounding
$-2.010 < x < -1.990$
So, I get $\delta = .01$
Would this be the answer to the question?
(It's my first question, please give constructive criticism)
 A: "I thought that by "suitable" it just means that I pick a random ϵ and then find the δ corresponding to it?"
No.
Consider this proof that $\lim_{x\rightarrow 64} \sqrt[6]{x} = 1.95$.  Let $\epsilon = .1$
Then $1.95 -.1 < \sqrt[6]{x} < 1.95 + .1$
So $1.85 < \sqrt[6] x < 2.05$
$40.089 < x < 74.22$ so let $\delta = 10$.  If $|x - 64| < 10$ then $|f(x) - 1.95| < .1$ so $\lim_{x\rightarrow 64} \sqrt[6]{x} = 1.95$.
This is obviously wrong because we didn't pick a small enough $\epsilon$.  We must prove this for all epsilon.
So
$-11 - \epsilon < 2x^2 + 9x - 1 < -11 + \epsilon$ 
$-10 - \epsilon < 2x^2 + 9x < -10 + \epsilon$
$-5 - \epsilon/2 < x^2 + \frac 92x < -5 + \epsilon/2$
$.0625 - \epsilon/2 < x^2 + \frac 92x + (\frac 94)^2 < .0625 + \epsilon/2$
$ 0\le x + \frac 94 < \sqrt {.0625 + \epsilon/2}$
$0\le x + \frac 94 < \sqrt {.0625 + \epsilon/2} < \sqrt {.0625} + \sqrt{\epsilon/2}$
$0\le x + 2.25 < .25 + \sqrt{\epsilon/2}$
$-2 - \sqrt{\epsilon/2} < -2 \le x < -2 + \sqrt{\epsilon/2}$
So let $\delta = \sqrt{\epsilon/2}$ for ANY $\epsilon$ and then $|x - (-2)| < \delta = \sqrt {\epsilon/2}$ will imply $|2x^2 + 9x - 1 -(-11)| < \epsilon$
