[Epsilon-Detla] Finding Epsilon

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bbl

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I'm familiar with proving using Epsilon-Delta but I got stumped with this problem because there were many left-overs (such as x-1, x+1, x^2 + 4), so how will I go about finding the epsilon here? Thank you in advance!

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I believe you're referring to finding delta in terms of epsilon in the scratch work.
[math]|f(x)-L|<\epsilon\\ \left|\frac{x^2+1}{x^2+4}-\frac{2}{5}\right|=\left|\frac{3(x-1)(x+1)}{5(x^2+4)}\right|=|x-1|\cdot \left|\frac{3}{5}\right|\cdot\frac{|x+1|}{x^2+4}<\epsilon[/math].
Put a constraint on [imath]\delta[/imath]. Suppose [imath]\delta <1 [/imath], then [imath]|x-1|<1 \implies[/imath][imath]-1< x-1<1\implies 0< x < 2[/imath].
Let's examine [imath]\frac{|x+1|}{x^2+4}[/imath]. The largest numerator occurs when [imath]x=2[/imath] i.e. [imath]2+1=3[/imath]. On the other hand, the smallest the denominator can be is when [imath]x=0[/imath] i.e. [imath]0^2+4=4[/imath]. So,
[math]\frac{|x+1|}{x^2+4}\le\frac{3}{4}[/math]It follows that:
[math]|x-1|\cdot \left|\frac{3}{5}\right|\cdot\frac{|x+1|}{x^2+4} \le|x-1|\cdot \left|\frac{3}{5}\right|\cdot\frac{3}{4}=\frac{9}{4}|x-1|<\epsilon[/math].
Therefore, we choose [math]\delta<\min\left\{1,\frac{4\epsilon}{9}\right\}\quad \text{or}\quad \delta=\frac{1}{2}\left\{1,\frac{4\epsilon}{9}\right\}[/math]If someone can confirm my argument.
 
BigBeachBanana said:
I believe you're referring to finding delta in terms of epsilon in the scratch work.
[math]|f(x)-L|<\epsilon\\ \left|\frac{x^2+1}{x^2+4}-\frac{2}{5}\right|=\left|\frac{3(x-1)(x+1)}{5(x^2+4)}\right|=|x-1|\cdot \left|\frac{3}{5}\right|\cdot\frac{|x+1|}{x^2+4}<\epsilon[/math].
Put a constraint on [imath]\delta[/imath]. Suppose [imath]\delta <1 [/imath], then [imath]|x-1|<1 \implies[/imath][imath]-1< x-1<1\implies 0< x < 2[/imath].
Let's examine [imath]\frac{|x+1|}{x^2+4}[/imath]. The largest numerator occurs when [imath]x=2[/imath] i.e. [imath]2+1=3[/imath]. On the other hand, the smallest the denominator can be is when [imath]x=0[/imath] i.e. [imath]0^2+4=4[/imath]. So,
[math]\frac{|x+1|}{x^2+4}\le\frac{3}{4}[/math]It follows that:
[math]|x-1|\cdot \left|\frac{3}{5}\right|\cdot\frac{|x+1|}{x^2+4} \le|x-1|\cdot \left|\frac{3}{5}\right|\cdot\frac{3}{4}=\frac{9}{4}|x-1|<\epsilon[/math].
Therefore, we choose [math]\delta<\min\left\{1,\frac{4\epsilon}{9}\right\}\quad \text{or}\quad \delta=\frac{1}{2}\left\{1,\frac{4\epsilon}{9}\right\}[/math]If someone can confirm my argument.
Click to expand...
If I may ask, where did you get [math]\delta = \frac{1}{2}[/math]?
 
bbl said:
If I may ask, where did you get [math]\delta = \frac{1}{2}[/math]?
Click to expand...
My mistake. I meant to write:
[math]\delta<\min\left\{1,\frac{4\epsilon}{9}\right\}\quad \text{or}\quad \delta=\frac{1}{2}\red{\min}\left\{1,\frac{4\epsilon}{9}\right\}[/math]One-half was an arbitray choice to turn the inequality into equality.
 
BigBeachBanana said:
My mistake. I meant to write:
[math]\delta<\min\left\{1,\frac{4\epsilon}{9}\right\}\quad \text{or}\quad \delta=\frac{1}{2}\red{\min}\left\{1,\frac{4\epsilon}{9}\right\}[/math]One-half was an arbitray choice to turn the inequality into equality.
Click to expand...
I see! Thank you very much! This helped a lot :)
 
Here is how i do these type problem--very close to BBB methods.
Choose |x-1|<δ <1
That is -1<x-1<1
Add 2 to both sides to get x+1 in the middle
1<x+1<3 => |x+1| < 3

Now for x^2 + 4.
If -1<x-1<1, then 0 <x<2.
Then 0<x^2<4
Now 4 < x^2 + 4 <8 and 1/8 <1/(x^2 + 4) < 1/4
Therefore 1/|x^2 + 4| < 1/4

Putting all this together we get that |x+1|/(x^2 + 4)ϵ < (3/4)ϵ
So (3/5)|x+1|/(x^2 + 4)ϵ < (3/5)(3/4)ϵ = 9ϵ/20

In the end, choose δ = min{ 9ϵ/20, 1}

Note that choosing δ<1 is completely arbitrary.

I haven't given it any real thought by I wonder if BBB is correct in saying that δ < min{ 9ϵ/20, 1} (???)
 
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Steven G said:
In the end, choose δ = min{ 9ϵ/20, 1}

Note that choosing δ<1 is completely arbitrary.

I haven't given it any real thought by I wonder if BBB is correct in saying that δ < min{ 9ϵ/20, 1} (???)
Click to expand...
First, I want to point out that [imath]\delta <\min\left\{1,\frac{20\epsilon}{9}\right\}[/imath].
The reason I say "less than" is because we restricted [imath]\delta <1[/imath].
 
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