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Epsilon Delta Definition of a Limit.md

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+#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](https://www.desmos.com/calculator/tucchymbrq)