See also: Proving Limits of Square Roots

# Basic Structure

For example, if you want to solve the limit below within [100, 108], you need a range within x = [9.79, 10.198]. You want to show that you can make the y-window [102, 106] $\to$ [103,105] $\to$ [103.99,104.01], and there will always exist an x-window.

# Proving Quadratic Limits

## Discovery phase

**never**appear in an actual proof. Think of it like your scratch paper. You would never turn that in as the actual answer. In this step, we're just finding a delta for a given epsilon.

Instead of starting with a delta, you start with stating what the epsilon is. $$|f(x) - L| < \epsilon$$ This means the function is within $\epsilon$ of the limit. $$|x^2 + 4 - 104| < \epsilon$$ $$|x^2 - 100| < \epsilon$$ $$|x-10||x+10| < \epsilon$$ Now, we arrive at a problem. Let's say we write this: $$|x-10| < \dfrac{\epsilon}{|x + 10|}$$ If we pick an arbitrary value for $\epsilon$, say 10, what is our $\delta$? We don't know from the equation above.

So, we have to say this. Let $\delta$ be at most 1. That way, for really big values of $\epsilon$, $\delta$ = 1 is still correct, while for small values of $\epsilon$, $\delta$ is unchanged. So, x $\in$ [9, 11] $$9|x-10| \leq |x+10||x-10| \leq 11|x-10|$$ Now, if we let $11|x-10| < \epsilon$, then that means that $|x+10||x-10| < \epsilon$

Now, we've arrived at the end of our discovery phase: $$|x-10| < \dfrac{\epsilon}{11}$$

# Real Proof

David Witten