How Does Calculating the Limit Help Find g in Advanced Mathematics?

[JasonRox]

How do I find g?

It's so confusing.

I'm trying to learn this on my own, so bare with me.

I'm going with an example I know the answer to, and maybe someone can work with me here. I'll ask questions through the solution.

We will do x^3 since that is complicated enough, but I understand the steps, just not the logic to moving on to the next step.

$$|x^3-a^3|<e$$
$$0<|x-a|<g$$

Find a value for g that satisfies the above.

$$|x^3-a^3|=|x-a||x^2+ax+a^2|$$
$$|x^2+ax+a^2|<=|x|^2+|a||x|+|a|^2$$

Note: $$|x|-|a|<=|x-a|<1$$
If you don't understand why one is chosen maybe this isn't for you. In case you forgot, we take 1 because that guarantees that the difference won't be too big.

$$|x|<1+|a|$$

Take the above and you get...

$$(1+|a|)^2+|a|(1+|a|)+|a|^2<br /> <br /> WARNING: This is in the works. I will be back to complete it.$$

 

[matt grime]

You'd be better using that |a|< |x| +g

or better yet, assuming that g is chosen such that |a|<2|x|, since if some g works, a smaller g has to work too, so there's no harm in placing a maximal size on g that helps eliminate a (assuming x is not zero. if x is zero it's quite easy)

 

[M98Ranger]

The precise definition of a limit is a mathematical concept that describes the behavior of a function as the input approaches a certain value. In other words, it determines what value a function is approaching as the input gets closer and closer to a specific value. To find the limit, we use a specific notation: lim f(x) = L as x approaches a, which means that the limit of f(x) is L as x gets closer and closer to a.

To find g in the given problem, we need to find a value for g that satisfies the inequalities |x^3-a^3|<e and 0<|x-a|<g. This means that the difference between x^3 and a^3 must be smaller than e, and the difference between x and a must be smaller than g.

To solve for g, we can start by simplifying the expression |x^3-a^3|. Using the formula for the difference of cubes, we can rewrite it as |x-a||x^2+ax+a^2|. Now, we want to make this expression smaller than e, so we can set up the inequality |x^2+ax+a^2|<=e.

Next, we can use the triangle inequality to simplify the expression further. This states that |x+y|<=|x|+|y| for any numbers x and y. In our case, we can rewrite |x^2+ax+a^2| as |x|^2+|a||x|+|a|^2.

Now, we can use the given inequality 0<|x-a|<g to solve for g. We know that |x|-|a|<=|x-a|<1, so we can substitute |x|-|a| for |x-a| in the inequality |x|^2+|a||x|+|a|^2<=e. This gives us (|x|-|a|)^2+|a|(|x|-|a|)+|a|^2<=e.

To make this expression even smaller, we can choose a value for |x| that is close to |a|. In fact, if we choose |x|<1+|a|, we can guarantee that the expression will be smaller than e. This is because (1+|a|)^2+|a|(1+|a|)+|a|^2 is