00:00 Intro

00:40 Epsilon-Delta criterion

04:32 Proof

08:28 Credits

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 $|f(x) - f(x_0)| \geq \varepsilon$.