Proving the limit of a quadratic

See also: Proving Limits of Square Roots

Basic Structure

The method we will use to prove the limit of a quadratic is called an epsilon-delta proof. The basic idea of an epsilon-delta proof is that for every y-window around the limit you set, called epsilon ($\epsilon$), there exists an x-window around the point, called delta ($\delta$), such that if x is in the x-window, f(x) is in the y-window.

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

$$\lim_{x \to 10}\left(x^2 + 4\right) = 104$$

Discovery phase

This is something I call the "discovery phase." This will 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

We want to prove this statement: $$\forall \epsilon > 0, \exists \delta \text{ }s.t. \text{ if } 0 < |x-10| < \delta \to |x^2 - 100| < \epsilon$$ In order to prove a $\forall$ statement, you have to start with an arbitrary value. That way, if you show it works for that arbitrary value, it will work for any value. $$\text{Let } \epsilon > 0$$ Now, there are two cases: one where $\epsilon > 11$ and another where $\epsilon < 11$. That's because $\epsilon/11$ is either $\gt 1$ or $\leq 1$ $$\text{Case 1: } \epsilon > 11$$ $$\text{Let }\delta = \min(1, \dfrac{\epsilon}{11}) = 1 \text{ because } \dfrac{\epsilon}{11} > 1$$ $$|x-10|< 1 \to 11|x-10| < 11$$ $$|x+10||x-10| \leq 11|x-10| < 11 < \epsilon$$ $$|x+10||x-10| < \epsilon$$ $$\text{ Now, we're done with this case. Let's move on to the next one.}$$ $$\text{Case 2: } \epsilon > 0 \text{ and } \epsilon \leq 11$$ $$\text{Let }\delta = \min(1, \dfrac{\epsilon}{11}) = \dfrac{\epsilon}{11} \text{ because } \dfrac{\epsilon}{11} \leq 1$$ $$|x-10|< \dfrac{\epsilon}{11} \to 11|x-10| < \epsilon$$ $$|x+10||x-10| \leq 11|x-10| < 11 < \epsilon$$ $$|x+10||x-10| < \epsilon$$ We just showed that for an arbitrary $\epsilon > 0$, there exists a $\delta$ such that if $|x-10| < \delta$, then $|x+10||x-10| < \epsilon$. Pretty much, we just did the discovery proof backwards.