1. ## Delta Epsilon Proof

Hello, I read the stickied delta epsilon guide at the top of the page and found it very helpful for finding an epsilon for a given delta. The guide represents delta = epsilon/M. If M has an x in it, we need to set a bound on delta in order to get an M that is a number without any x's in it.

My problem is, for example, proving that the limit of x^3 as x approaches 3 is 27. Here is what I have so far:

$|x-3| < \delta$

$|x^3 - 27| < \epsilon$

$(|x - 3|)(|x^2 + 3x + 9|) < \epsilon$

$(|x -3|) < \epsilon/(|x^2 + 3x + 9|)$

In the stickied article, it says that I need to set a delta (i.e, delta = 1) and then add things to my first equation so that $|x - 3|$ becomes $|x^2 + 3x + 9|$. If i do this, my M still ends up having an x in it because the only way to turn |x -3| into a quadratic equation is to multiply by x. I can solve delta epsilon proofs, but the cubic powers are throwing me off here. Any suggestions?

2. If $\delta <1$ then $\left| {x - 3} \right| < \delta \, \Rightarrow \,\left| x \right| < 4$.

$\left| {x^2 + 3x + 9} \right| < 16 + 12 + 9$

3. Thanks for the help, Plato! I didn't realize it was that simple.

I have one more question that I was wondering if someone could look over.

I am trying to prove $e^x$ is continuous. Can someone please make sure I'm doing this right?

Scratch work:

$|x - x_0| < \delta$

$|e^x - e^x_0| < \epsilon$

$ln |e^x - e^x_0| < ln |\epsilon|$

$|x - x_0| < ln |\epsilon|$

$\delta < (ln |\epsilon|)$

Proof:

Let $\delta = min{(1, ln |\epsilon|)}$

$|x - x_0| < \delta$

$|e^x - e^x_0| < e^\delta$

$|e^x - e^x_0 |< e^(ln |\epsilon|)$ (ln |epsilon| should be in the exponent)

$|e^x - e^x_0| < \epsilon$

$|f(x) - f(x_0)| < \epsilon$, therefore $e^x$ is continuous.

4. Originally Posted by Epiphenomenon

$ln |e^x - e^x_0| < ln |\epsilon|$

$|x - x_0| < ln |\epsilon|$
you assumed that the logarithm is linear, so this is step is wrong.

5. Oh, ok, I will work on it. How would you recommend I approach this?

6. Ok, I can solve for delta, but I can't create an equation to substitute it into.

$x - x_0 < \delta$

$e^x - e^x_0 < \epsilon$

$e^x < \epsilon + e^x_0$

$ln|e^x| < ln |\epsilon + e^x_0|$

$x < ln |\epsilon + e^x_0|$

$x - x_0 < ln|\epsilon + e^x_0| - x_0$

$\delta < ln|\epsilon + e^x_0| - x_0$

Now that I have my delta written in terms of epsilon and x0, I can't figure out how to substitute it into $e^x - e^x_0$ to get it $< \epsilon$

7. Is it ok if I start with my delta equation $|x - x_0| < \delta$, make the necessary substitions into my delta equation, and end up with $|f(x) - f(x_0) < \epsilon$ for my proof? As opposed to starting with $|f(x) - f(x_0)|$ and ending with epsilon? Can I start my proof with the delta equation instead of the epsilon equation?

Here's what I'm thinking of doing:

$\delta = min(1, ln |\epsilon + e^x_0| - x_0)$

$x - x_0 < \delta$

$e^x/e^x_0 < e^\delta$ (I raised the delta equation by the power of e on both sides)

$e^x/e^x_0 < e^(ln |(\epsilon + e^x_0)| - x_0)$

$e^x/e^x_0 < (\epsilon + e^x_0)/e^x_0$

$e^x - e^x_0 < \epsilon$

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,

### prove that lim x~3 x^3=27

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