What are the mechanics of finding delta in a limit problem?

[Saladsamurai]

Okay Then! I am going to start with a simple problem here:

Given some function, a limit L, an x_o, and some \epsilon:

a) Find an open interval on which the inequality |f(x)-L|<\epsilon holds. Then b) give a value for \delta>0 such that
for all x satisfying 0 < |x - x₀| < \delta\Rightarrow |f(x)-L|&lt;\epsilon.

f(x)=x+1
L = 5
x_o=4
\epsilon=0.01

a) To find an interval on which |f(x)-L|&lt;\epsilon holds, I simply solve the inequality:

|f(x)-L|&lt;\epsilon

-\epsilon&lt;f(x)-L&lt;\epsilon

-\epsilon&lt;x+1-5&lt;\epsilon

-\epsilon&lt;x-4&lt;\epsilon

4-\epsilon&lt;x&lt;4+\epsilon

3.99&lt;x&lt;4.01

So there is my open interval, (3.99, 4.01), on which |f(x)-L|&lt;\epsilon holds.

Now I know that for part (b), delta must be 0.01.

But what how do we actually find \delta? What are the mechanics of finding it.

For part (a) I solved an inequality; what did I do for part (b) to find delta?

Sorry if this is a little vague, I am not sure exactly how to word my question.

 

[snipez90]

A better question to ask is what is delta? Delta is how sufficiently close x must be to x_0 to guarantee that |f(x) -L| < epsilon will be satisfied. Thus, we want |x-x_0| &lt; \delta or x_0 - \delta &lt; x &lt; x_0 + \delta. In this case, x_0 = 4 so we want a delta such that 4 -\delta &lt; x &lt; 4 + \delta.

Now compare this last set of inequalities to what you concluded in part a). How do we choose delta to ensure that |f(x) - L| is indeed less than epsilon?

 

[Saladsamurai]

A better question to ask is what is delta? Delta is how sufficiently close x must be to x_0 to guarantee that |f(x) -L| < epsilon will be satisfied. Thus, we want |x-x_0| &lt; \delta or x_0 - \delta &lt; x &lt; x_0 + \delta. In this case, x_0 = 4 so we want a delta such that 4 -\delta &lt; x &lt; 4 + \delta.

Now compare this last set of inequalities to what you concluded in part a). How do we choose delta to ensure that |f(x) - L| is indeed less than epsilon?

So since we know that we need |x-x_0| &lt; \delta (1), we also know x_o=4 (2), and we have an inequality that says 3.99 < x < 4.01 (3), we can simply 'combine' (1), (2), and (3) to yield ...err something. I need a moment to think about it. But, I think I see it now.

 

[Saladsamurai]

I am still a little lost here sorry. What is the next step?

I know that 3.99<x<4.01. I also know that |x-x_o|<\delta. And x_o=4.

How do I combine the 3 into something meaningful to find \delta ?

Thanks

 

[Saladsamurai]

Nevermind. I was forgetting to subtract 4 from ALL sides of the inequality.

Using the above we have:

3.99-4 < x-4 < 4.01-4

-0.01<x-4< 0.01
or
|x-4|<0.01 = delta