public-notes/Epsilon Delta Definition of a Limit.md

#Math #Calculus

Definition

When

\lim _{x \to c} f(x) = L

For \epsilon > 0 and \delta > 0, there is a value \delta for every value of \epsilon such that

0 < |x - c| < \delta\\
0 < |f(x) - L| < \epsilon\\

Proving a Limit

Let’s prove:

\lim _{h \to 0} \frac {(x + h)^2 - x^2} h = 2x

Let:

0 < |\frac {(x + h)^2 - x^2} h - 2x| < \epsilon \\
0 < |\frac {(x + h)^2 - x^2} h - 2x| < \epsilon \\0 < |\frac {(x^2 + 2xh + h^2 - x^2)} h - 2x| < \epsilon \\0 < |\frac {2xh + h^2} h - 2x| < \epsilon \\

Remember \epsilon > 0:

0 < |2x + h - 2x| < \epsilon \\
0 < |h| < \epsilon

We have to prove for every \epsilon:

0 < |h - 0| < \delta \\
0 < |h| < \delta

These two inequalities are the same, so they are easily satisfied just by setting:

\delta = \epsilon

Graphical Explanation

https://www.desmos.com/calculator/tucchymbrq