MHB Tough limit proof for a general f(x) and g(x)

[Amad27]

Hello,

$f(x)$ and $g(x)$ have the following property: for all $\epsilon > 0$ and all $x$,
$$ \text{if} \space 0 < |x - 2| < \sin^2(\epsilon^2/9) + \epsilon \space \text{then} \space |f(x) - 2| < \epsilon$$
$$ \text{if} \space 0 < |x - 2| < \epsilon^2 \space \text{then} \space |g(x) - 4| < \epsilon$$

Find a $\delta$ such that:
(i) if $0 < |x-2| < \delta \space \text{then} \space |f(x) + g(x) - 6| < \epsilon$

This problem is from Spivak's calculus, so no wonder is it quite hard. I understand a few things.

The trick is to get $|f(x) - 2| < \epsilon/2$ and $|g(x) - 2| < \epsilon/2$

I need some help, as it is very tough to find a $\delta$ that will do. First should we bound $|x - 2| < c$ for some integer (positive) $c$?

 

[Fantini]

Here's how I approached this problem. Naming

$$\begin{aligned} \delta_1(\varepsilon) & = \sin^2 \left( \frac{\varepsilon^2}{9} \right) + \varepsilon, \\ \delta_2 (\varepsilon) & = \varepsilon^2, \end{aligned}$$

we have that if $0 < |x-2| < \delta_1 \left( \frac{\varepsilon}{2} \right)$ then $|f(x) -2 | < \frac{\varepsilon}{2}$. The same logic applies to $\delta_2 \left( \frac{\varepsilon}{2} \right)$. Computing them gives us

$$\begin{aligned} \delta_1 \left( \frac{\varepsilon}{2} \right) & = \sin^2 \left( \frac{\varepsilon^2}{36} \right) + \frac{\varepsilon}{2}, \\ \delta_2 \left(\frac{\varepsilon}{2} \right) & = \frac{\varepsilon^2}{4}. \end{aligned}$$

Taking the minimum

$$\delta = \min \left\{ \delta_1 \left( \frac{\varepsilon}{2} \right), \delta_2 \left(\frac{\varepsilon}{2} \right) \right\} = \min \left\{ \sin^2 \left( \frac{\varepsilon^2}{36} \right) + \frac{\varepsilon}{2}, \frac{\varepsilon^2}{4} \right\}$$

ensures that $0 < |x-2| < \delta$ guarantees $|f(x) - 2 | < \frac{\varepsilon}{2}$ and $|g(x) - 4 | < \frac{\varepsilon}{2}$.

Therefore $0 < |x-2| < \delta$ implies that

$$|f(x) + g(x) -6| \leq |f(x) - 2| + |g(x) -4| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$

But I'm not completely sure on this.

 

[Amad27]

Here's how I approached this problem. Naming

$$\begin{aligned} \delta_1(\varepsilon) & = \sin^2 \left( \frac{\varepsilon^2}{9} \right) + \varepsilon, \\ \delta_2 (\varepsilon) & = \varepsilon^2, \end{aligned}$$

we have that if $0 < |x-2| < \delta_1 \left( \frac{\varepsilon}{2} \right)$ then $|f(x) -2 | < \frac{\varepsilon}{2}$. The same logic applies to $\delta_2 \left( \frac{\varepsilon}{2} \right)$. Computing them gives us

$$\begin{aligned} \delta_1 \left( \frac{\varepsilon}{2} \right) & = \sin^2 \left( \frac{\varepsilon^2}{36} \right) + \frac{\varepsilon}{2}, \\ \delta_2 \left(\frac{\varepsilon}{2} \right) & = \frac{\varepsilon^2}{4}. \end{aligned}$$

Taking the minimum

$$\delta = \min \left\{ \delta_1 \left( \frac{\varepsilon}{2} \right), \delta_2 \left(\frac{\varepsilon}{2} \right) \right\} = \min \left\{ \sin^2 \left( \frac{\varepsilon^2}{36} \right) + \frac{\varepsilon}{2}, \frac{\varepsilon^2}{4} \right\}$$

ensures that $0 < |x-2| < \delta$ guarantees $|f(x) - 2 | < \frac{\varepsilon}{2}$ and $|g(x) - 4 | < \frac{\varepsilon}{2}$.

Therefore $0 < |x-2| < \delta$ implies that

$$|f(x) + g(x) -6| \leq |f(x) - 2| + |g(x) -4| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$

But I'm not completely sure on this.

$$\delta_1(\epsilon) = \sin^2(\epsilon^2/9) + \epsilon$$How does the statement imply:

$0 < |x - 2| < \delta_1(\epsilon/2)$ then $|f(x) - 2| < \epsilon/2$?

 

[Fantini]

I'm sorry, I didn't understand. How does what statement imply what? :(

 

[Amad27]

I'm sorry, I didn't understand. How does what statement imply what? :(

Sorry for being unclear! I meant from:

$0 < |x - 2| < \delta_1(\epsilon/2)$

How do you get:

$|f(x) - 2| < \epsilon/2$?

 

[Fantini]

You compute $\delta_1(\varepsilon)$ so that $|f(x)-2|< \varepsilon$. From here you know that if you want $|f(x)-2| < h(\varepsilon)$ then you need to have $\delta_1(h(\varepsilon))$. You have a correspondence between the argument of $\delta_1(h(\varepsilon))$ and what $|f(x)-2|$ is less than. :)