# Homework Help: Epsilons and Deltas and Such

1. Sep 12, 2009

### Saladsamurai

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

Given some function, a limit L, an xo, 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| < $\delta\Rightarrow |f(x)-L|<\epsilon$.

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

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

$$|f(x)-L|<\epsilon$$

$$-\epsilon<f(x)-L<\epsilon$$

$$-\epsilon<x+1-5<\epsilon$$

$$-\epsilon<x-4<\epsilon$$

$$4-\epsilon<x<4+\epsilon$$

$$3.99<x<4.01$$

So there is my open interval, (3.99, 4.01), on which $|f(x)-L|<\epsilon$ 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.

2. Sep 12, 2009

### 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 $|x-x_0| < \delta$ or $x_0 - \delta < x < x_0 + \delta.$ In this case, $x_0 = 4$ so we want a delta such that $4 -\delta < x < 4 + \delta$.

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?

3. Sep 12, 2009

### Saladsamurai

So since we know that we need $|x-x_0| < \delta$ (1), we also know xo=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.

4. Sep 12, 2009

### 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-xo|<$\delta$. And xo=4.

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

Thanks

5. Sep 12, 2009

### 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

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