is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Proving the limit of a quadratic

Basic Structure

MathJax TeX Test Page 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 ($\delta$).

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

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

Discovery phase

MathJax TeX Test Page Instead of starting with an epsilon, you start with stating what the delta is. $$0 < |x-10| < \delta$$ This means that the x you choose is closer than $\delta$ to 10. Now, we start with the work: $$|f(x) - L| < \epsilon$$ that means the function is less than $\epsilon$ away from the limit. $$|x^2 + 4 - 104| < \epsilon$$ $$|x^2 - 100| < \epsilon$$ $$|x-10||x+10| < \epsilon$$ Now, we arrive a problem, because recall what you're trying to do here: find a relationship between $\delta$ and $\epsilon$, and you can't do that using the relationship $$|x-10| < \delta, |x-10| < \frac{\epsilon}{|x+10|} \to \delta = \frac{\epsilon}{|x+10|}$$ If you were given $\epsilon = 10$, you wouldn't know what $δ$ is, so we know that doesn't current expression work. Now, we can do another inequality by setting delta to $2$, so x is in the range $[8, 12]$. $$8|x-10| ≤ |x+10||x-10| ≤ 12|x-10|$$ Now, if we prove that $12|x-10| < \epsilon$, then we prove that $|x+10||x-10| < \epsilon$ So, our goal is to now prove this: $$|x-10| < \dfrac{\epsilon}{12}$$ We now let $\delta = \min(\dfrac{\epsilon}{12}, 2)$ The reason you do that is you assumed that $\delta$ was at most 2 before, so now you have to set a bound that $\delta$ is the minimum of $\frac{\epsilon}{12}$ and 2. $$|x-10| < \delta = \dfrac{ε}{12}$$ $$12|x-10| < \epsilon $$ $$|x+10||x-10| ≤ 12|x-10| < \epsilon$$ $$|x^2 - 100| < \epsilon$$ $$|x^2 + 4 - 104| < \epsilon$$ $$|f(x) - L| < \epsilon$$ Now, we proved it! So we just did the the discovery proof backward, and we proved that $$\lim_{x \to 10}x^2 + 4 = 104$$
David Witten

Implicit Differentiation

Finding the Integral of sec(x)