How do you find delta in an epsilon-delta proof of continuity when the function is nonlinear and can't be factored? So we define $f:\mathbb{R}\to \mathbb{R}$ and $f(x)=x^2+3x-5$, and we're using the epsilon-delta definition to prove that it is continuous at 3.
So I know the general structure of an epsilon-delta proof, but we were never shown how to prove a nonlinear function. Like I have no idea what to choose delta as, and how to even figure out what we should choose delta as. I know we need to let epsilon be an arbitrary positive real number, and plug the equation so that $|x^2+3x-5-3|<\delta$ so that $|x^2+3x-8|<\delta$. I don't think you can factor this, so what do you do in order to choose delta?
 A: You have that $f(3)=3^2+3\cdot 3-5=13$.
This leads to $|x^2+3x-5-13|=|x^2+3x-18|=|x-3||x+6|$.
Under the assumption that $|x-3|<\delta$ we now have to stipulate $|x+6|$.
The trick is to stipulate. Take some $\delta$. Lets say $\delta=1$ (the value does not matter).
Then $|x-3|<1\Leftrightarrow x-3<1<x+3$. Then $x+6<10$, by just adding 9 to the inequality.
So $|x+6|<10$ when $\delta<1$.
We then get $|x-3||x+6|<10\delta$.
So we have to choose $\delta=\min\{1,\frac{\varepsilon}{10}\}$ to conclude the proof.
A: It is often the case that the function is monotonous in some neighborhood of the point considered, so that the inequation
$$-\epsilon<f(x)-f(a)<\epsilon$$ can be solved by
$$f^{-1}(f(a)-\epsilon)<x<f^{-1}(f(a)+\epsilon)$$ (or bounds swapped if the function is decreasing).
From this,
$$f^{-1}(f(a)-\epsilon)-a<x-a<f^{-1}(f(a)+\epsilon)-a$$  and
$$|x-a|<\min(a-f^{-1}(f(a)-\epsilon),f^{-1}(f(a)+\epsilon)-a).$$
The RHS gives an upper bound for $\delta$.

In more difficult cases, you have to figure out the inverse image of $(f(a)-\epsilon,f(a)+\epsilon)$ and intersect it with an interval of width $2\delta$ centered at $a$.
In practice, it is possible to find shortcuts and gross estimates that fulfill the conditions. Solving the inequalities need not be done in a tight way.
For example, for the limit of $x^2+3x-5$ at $x=3$, we have to achieve $|x^2+3x-18|<\epsilon$. As the slope of the curve is $9$ at the target point, we can try a smaller mutiple, say $\delta=\dfrac{\epsilon}{10}$.
We evaluate
$$\left(3\pm\frac\epsilon{10}\right)^2+3\left(3\pm\frac\epsilon{10}\right)-18=\left(\frac{\epsilon}{100}\pm\frac{9}{10}\right)\epsilon,$$ which is smaller than $\epsilon$ for all $\epsilon<10$.
A: Let $\delta$ be the value we want and lets try to find what restrictions we need.
$|x - 3| < \delta$
$3 -\delta < x < 3 + \delta$.  As $\delta$ can be arbitrarily small we can assume will always pick a $\delta \le 3$.
$0\le 9-6\delta + \delta^2 < x^2 < 9 +6\delta + \delta^2$
$(9 - 6\delta + \delta^2) + 3(3-\delta) -5 < x^2 + 3x -5 < (9 + 6\delta + \delta^2) + 3(3+\delta) - 5$
$13 - 9\delta + \delta^2 < x^2 +3x-5 < 13 +9\delta + \delta^2$
$-9\delta + \delta^2 < (x^2 + 3x - 5) - 13 < 9\delta + \delta^2$.
now as $\delta$ can be arbitrarily small we can assume we will always pick a $\delta \le 1$.  And if $0 < \delta \le 1$ then $\delta^2 \le \delta$.
$9\delta + \delta^2 \le 9\delta + \delta = 10\delta$.  And $-9\delta + \delta^2 >-9\delta > -10\delta$.
So
$-10\delta \le -9\delta + \delta^2 < (x^2 + 3x - 5)-13 <9\delta + \delta^2 < 10 \delta$.
SO
$-10\delta  < (x^2 + 3x - 5)-13  < 10 \delta$
$|(x^2 + 3x - 5)-13| < 10\delta$.
$|(x^2 + 3x -5) - (3^3 + 3\cdot 3 - 5)| < 10\delta$
So if we choose a $\delta$ where $\delta \le \frac \epsilon {10}$ and $\delta \le 1$ and $\delta \le 3$ we will be good and proven our result.
So let $\delta = \min(\frac {\epsilon}{10}, 1)$.
A: Another way of looking at it is to rewrite your polynomial as one centered at $x-3$
$$(x-3)^2=x^2-6x+9$$, we are off by $9x$, so we add $9(x-3)$ to get
$(x-3)^2 +9(x-3)=x^2+3x-18$.  This is off by 13, so we add 13 to get
$$(x-3)^2+9(x-3)+13= x^2+3x-5$$,  your polynomial.  Now you get to see
$$|f(x)-f(3)|=|(x-3)^2 +9(x-3)+13-13|=|(x-3)^2+9(x-3)|$$
The advantage of looking at it like this is we can consider how small do we need |x-3| to be to get this under any arbitrary $\epsilon$.  Using the triangle inequality,
$$|(x-3)^2+9(x-3)|\leq |(x-3)^2|+9|x-3|$$
For numbers where $|a|\leq 1$, squaring preserves the order,  so $|x-3|^2\leq |x-3|$ as long as $\delta \leq 1$, and since we get to pick $\delta$, we can make this happen.  This simplifies our inequality to
$$|f(x)-f(3)|\leq |x-3|+9|x-3|=10|x-3|$$
so as long as $|x-3|\leq \frac \epsilon {10}$  and 1 at the same time,  we are good,  so we take the minimum of 1 or $\frac \epsilon {10}$
The general idea here was to reframe your limit as a limit at 0 instead of a limit at 3.  Then we can show how close do we need the inputs to be from 0 to keep the outputs close to 0.  This works well for functions that near 0 are small,  like polynomials.
