# 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 $\delta,\epsilon$ proof is to show $|x - a| < \delta \implies |f(x) - f(a) | < \epsilon$. You do not need, use, want, or have $|f(x) -a| < \delta$ which is what you are attempting to use. $(x^2 + 3x-5)$ is $f(x)$ . And $3$ is the $x = a$ value. You don't combine those in any way. You want $|\color{blue}x -3|< \delta \implies |x^2 + 3x -5 -\color{red}{13}|< \epsilon$. And $x^2+3x -5 -13 = x^2 + 3x -18$ which DOES factor if you want to do it by factoring. (You can see my asnwer to see how to avoid factoring altoghether.) – fleablood Apr 14 at 19:05

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 $$|x-3|<1\Leftrightarrow x-3<1. Then $$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.

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 $$\delta = \min(\frac {\epsilon}{10}, 1)$$.

It is often the case that the function is monotonous in some neighborhood of the point considered, so that the inequation

$$-\epsilon can be solved by

$$f^{-1}(f(a)-\epsilon) (or bounds swapped if the function is decreasing).

From this,

$$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 $$\epsilon<10$$.

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.