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Title: Epsilon-Delta Definition
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Series: Real Analysis
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Chapter: Continuous Functions
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YouTube-Title: Real Analysis 28 | Epsilon-Delta Definition
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Bright video: https://youtu.be/lWsy3Z8C1V0
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Dark video: https://youtu.be/EIYoxc2rlBs
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Ad-free video: Watch Vimeo video
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Original video for YT-Members (bright): https://youtu.be/4xhyqdjmxHU
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Original video for YT-Members (dark): https://youtu.be/UcFma9YNfaY
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Quiz: Test your knowledge
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Dark-PDF: Download PDF version of the dark video
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Print-PDF: Download printable PDF version
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Exercise Download PDF sheets
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Thumbnail (bright): Download PNG
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Thumbnail (dark): Download PNG
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Subtitle on GitHub: ra28_sub_eng.srt missing
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Timestamps
00:00 Intro
00:40 Epsilon-Delta criterion
04:32 Proof
08:28 Credits
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Subtitle in English (n/a)
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Quiz Content
Q1: Which of the following statements is equivalent to the statement: $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous at $x_0$.
A1: For all $\varepsilon > 0$ there is $\delta > 0$ such that for all $ x \in \mathbb{R}$ with the property $|x - x_0| < \varepsilon $ we also have $|f(x) - f(x_0)| < \delta$.
A2: For all $\varepsilon > 0$ there is $\delta > 0$ such that for all $ x \in \mathbb{R}$ with the property $|x - x_0| < \delta $ we also have $|f(x) - f(x_0)| < \varepsilon$.
A3: For all $\varepsilon > 0$ there is $\delta > 0$ such that for all $ x \in \mathbb{R}$ with the property $|f(x) - f(x_0)| < \delta $ we also have $|x- x_0| < \varepsilon$.
A4: For all $\varepsilon > 0$ there is $\delta > 0$ such that for all $ x \in \mathbb{R}$ with the property $|f(x) - f(x_0)| < \varepsilon $ we also have $|x- x_0| < \delta $.
Q2: The function $ \displaystyle f(x) = \begin{cases} 1 &, ~~ x > 0 \ 0 &, ~~x \leq 0\end{cases} ~$ is not continuous at $x_0 = 0$. For which of the following $\varepsilon$ we cannot find a number $\delta>0$ such that the $\varepsilon$-$\delta$-criterion is satisfied.
A1: $\varepsilon = \frac{1}{2}$
A2: $\varepsilon = 2$
A3: $\varepsilon = 3$
A4: $\varepsilon = 4$
Q3: Which of the following statements is equivalent to the statement: $f: \mathbb{R} \rightarrow \mathbb{R}$ is not continuous at $x_0$.
A1: There is $\varepsilon > 0$ such that for all $\delta > 0$ we find a number $ x \in \mathbb{R}$ with the property $|x - x_0| < \delta $ and $|f(x) - f(x_0)| \geq \varepsilon$.
A2: There is $\varepsilon > 0$ such that for all $\delta > 0$ we find a number $ x \in \mathbb{R}$ with the property $|x - x_0| < \varepsilon $ and $|f(x) - f(x_0)| \geq \delta$.
A3: There is $\varepsilon > 0$ such that there is $\delta > 0$ and we find a number $ x \in \mathbb{R}$ with the property $|x - x_0| < \delta $ and $|f(x) - f(x_0)| \geq \varepsilon$.
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Last update: 2025-01
