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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?

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  • $\begingroup$ 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.) $\endgroup$ – fleablood Apr 14 at 19:05
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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.

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

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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 $\epsilon<10$.

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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.

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