• Title: Epsilon-Delta Definition

  • Series: Real Analysis

  • Chapter: Continuous Functions

  • YouTube-Title: Real Analysis 28 | Epsilon-Delta Definition

  • Bright video: https://youtu.be/4xhyqdjmxHU

  • Dark video: https://youtu.be/UcFma9YNfaY

  • Quiz: Test your knowledge

  • PDF: Download PDF version of the bright video

  • Dark-PDF: Download PDF version of the dark video

  • Print-PDF: Download printable PDF version

  • Thumbnail (bright): Download PNG

  • Thumbnail (dark): Download PNG

  • Subtitle on GitHub: ra28_sub_eng.srt missing

  • Timestamps

    00:00 Intro

    00:40 Epsilon-Delta criterion

    04:32 Proof

    08:28 Credits

  • Subtitle in English (n/a)
  • 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 $|f(x) - f(x_0)| \geq \varepsilon$.

  • Back to overview page