55 lines
935 B
Markdown
55 lines
935 B
Markdown
#Math #Calculus
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# Definition
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When
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$$
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\lim _{x \to c} f(x) = L
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$$
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For $\epsilon > 0$ and $\delta > 0$, there is a value $\delta$ for every value of $\epsilon$ such that
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$$
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0 < |x - c| < \delta\\
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0 < |f(x) - L| < \epsilon\\
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$$
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# Proving a Limit
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Let’s prove:
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$$
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\lim _{h \to 0} \frac {(x + h)^2 - x^2} h = 2x
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$$
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Let:
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$$
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0 < |\frac {(x + h)^2 - x^2} h - 2x| < \epsilon \\
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0 < |\frac {(x + h)^2 - x^2} h - 2x| < \epsilon \\0 < |\frac {(x^2 + 2xh + h^2 - x^2)} h - 2x| < \epsilon \\0 < |\frac {2xh + h^2} h - 2x| < \epsilon \\
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$$
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Remember $\epsilon > 0$:
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$$
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0 < |2x + h - 2x| < \epsilon \\
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0 < |h| < \epsilon
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$$
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We have to prove for every $\epsilon$:
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$$
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0 < |h - 0| < \delta \\
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0 < |h| < \delta
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$$
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These two inequalities are the same, so they are easily satisfied just by setting:
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$$
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\delta = \epsilon
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$$
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# Graphical Explanation
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[https://www.desmos.com/calculator/tucchymbrq](https://www.desmos.com/calculator/tucchymbrq) |