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The formal definition of the limit is that $\lim_{x\to c} f(x)=L$ if and only if for any $\varepsilon >0$ there exists a $\delta >0$ such that $|x-c| <\delta \rightarrow |f(x)-L| < \varepsilon$. Use this definition for $\lim_{x \to 5} (x^2 -15x+50)$ to find the largest value for delta that satisfies epsilon equal to one.

Source: FAMAT State Convention 2019, Limits&Derivatives #29

Attempted solution: I set $|x^2-15x+50 - 0| < 1$ and solved for $x$, obtaining the following inequalities: $\frac{15-\sqrt{29}}{2} < x < \frac{15+\sqrt{29}}{2}$ and $x<\frac{15-\sqrt{21}}{2} $ or $ x > \frac{15+\sqrt{21}}{2}$ Next, I tried getting this in the form $|x-c| <\delta$ by changing each inequality as follows: $\frac{5-\sqrt{29}}{2} < x-5 < \frac{5+\sqrt{29}}{2}$ and $x-5<\frac{5-\sqrt{21}}{2} $ or $ x-5 > \frac{5+\sqrt{21}}{2}$

I don't know how to translate this result to $\delta$ because it does not follow a neat $-\delta<x-c<\delta \rightarrow |x-c| <\delta$

Answer provided by the competition:

$\frac{5-\sqrt{21}}{2}$

Edit: See comment to where I'm currently at

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  • $\begingroup$ Just pointing out that the question is flawed. The largest $\delta$ will result in $|f(x)-L|=\epsilon$. One needs to ask for the sup or lub of the set of $\delta$ that work. $\endgroup$ – Ted Shifrin Jul 25 '20 at 22:00
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I don't agree with the provided answer:

$|(x^2-15x+50) - 0| < 1$ is satisfied on two intervals: $(\frac{5-\sqrt{29}}{2}, \frac{5-\sqrt{21}}{2})$ and $(\frac{5+\sqrt{21}}{2}, \frac{5+\sqrt{29}}{2})$. Only the second interval contains $5.$ Now, $|\frac{5+\sqrt{21}}{2} - 5| = |\frac{-5+\sqrt{21}}{2}| = \frac{5-\sqrt{21}}{2} \approx 0.2087$ and $|\frac{5+\sqrt{29}}{2} - 5| = |\frac{-5+\sqrt{29}}{2}| = \frac{\sqrt{29}-5}{2} \approx 0.1926.$ We need to take the least of these as our largest possible $\delta,$ i.e. $\frac{\sqrt{29}-5}{2}.$

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  • $\begingroup$ $|(x^2-15x+50) - 0| < 1$ is satisfied on $(\frac{15-\sqrt{29}}{2}, \frac{15-\sqrt{21}}{2})$ and $(\frac{15+\sqrt{21}}{2}, \frac{15+\sqrt{29}}{2})$, of which only the first interval contains $5$.I turned this into an inequality and subtracted $5$ from each side to get $\frac{5-\sqrt{29}}{2}<x-5<\frac{5-\sqrt{21}}{2}$. If these are my two choices for $\delta$ that satisfy $\varepsilon = 1$, the largest is $\frac{5-\sqrt{21}}{2}$, though I'm not sure if I made a logic error somewhere in there. $\endgroup$ – physicsaficionado Jul 25 '20 at 23:03
  • $\begingroup$ @physicsaficionado. Oh, did I get the interval endpoints wrong? $\endgroup$ – md2perpe Jul 26 '20 at 17:09
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Let us cheat a little by using a plot.

Around $5$, we consider the roots of

$$x^2-15x+50=\pm1,$$

which are

$$5-\frac{\sqrt{29}-5}2$$ and $$5+\frac{5-\sqrt{21}}2.$$

The tightest is on the left, as you can see from the concavities.

enter image description here

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Wait! You know that when x= 5, $x^2- 15x+ 50= 0$? And when you solved for x you got $\frac{15\pm\sqrt{21}}{2}$? NO! Since $x^2- 15x+ 50= 0$ when x= 5 you know that x- 5 is a factor! And it is easy to see that $x- 10$ is the other.

$x^2- 15x+ 50= (x- 5)(x- 10)$ so $|x^2- 15x+ 50|< \epsilon$ gives |(x- 5)(x- 10)|= |x- 5||x- 10|< \epsilon. As long as x is not 10 we can write $|x- 5|< \epsilon/|x-10|$.

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