# Finding largest delta value given epsilon for delta-epsilon limit

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

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

• 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. – Ted Shifrin Jul 25 '20 at 22:00

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. $$\frac{\sqrt{29}-5}{2}.$$

• $|(x^2-15x+50) - 0| < 1$ 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. – physicsaficionado Jul 25 '20 at 23:03
• @physicsaficionado. Oh, did I get the interval endpoints wrong? – md2perpe Jul 26 '20 at 17:09

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. 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 $$|x- 5|< \epsilon/|x-10|$$.