Pointwise Convergence

• Explore pointwise convergence for the sequence $$f_n(x)=x^n \qquad (x\in[0,1]),$$ and its pointwise limit $$f(x)= \begin{cases} 0,& 0\le x<1,\\[4pt] 1,& x=1. \end{cases}$$
• For a fixed choice of $x$ and $\varepsilon>0$, the applet helps you see when $$|f_n(x)-f(x)|<\varepsilon \quad\text{for all } n \ge N_{x,\varepsilon}.$$
• Use the slider $n$ to choose the current index $n$ (the solid blue curve is $x^n$).
• Use the slider $\varepsilon$ to set the tolerance $\varepsilon>0$.
• Drag the point $x$ on the $x$-axis to choose the evaluation point $x\in[0,1]$.
• The solid blue curve is the current function $f_n(x)=x^n$.
• The faint dashed blue curves are earlier functions $x^k$ (for $k<n$), to show the trend as $n$ grows.
• The red limit function is drawn as:
  • the red line $y=0$ (limit on $[0,1)$),
  • the red point $(1,1)$ (since $1^n=1$ for all $n$).
• The vertical gray guide marks the chosen $x$.
• The dashed horizontal lines form the $\varepsilon$-tube around the limit value at the chosen $x$: $f(x)\pm\varepsilon$.
• The point $P_n=(x,f_n(x))$ shows the current value of the sequence at your chosen $x$.
• The dashed segment from $f(x)$ to $f_n(x)$ visualizes the error $|f_n(x)-f(x)|$.
• Fix an $x<1$ and increase $n$: the point $P_n=(x,x^n)$ moves down toward $0$, and eventually enters the $\varepsilon$-tube around $f(x)=0$.
• When $$|f_n(x)-f(x)|<\varepsilon,$$ the point $P_n$ (and the point $x$ on the axis) turn green.
• Drag $x$ closer to $1$ (e.g. $0.9,0.99,0.999$) while keeping $\varepsilon$ fixed: you will need much larger $n$ before $x^n<\varepsilon$. This shows that $N_{x,\varepsilon}$ depends strongly on $x$.
• Check the special case $x=1$: since $f_n(1)=1$ for all $n$, the limit value is $f(1)=1$ (the red point $(1,1)$), not $0$.

Description

Link to code.

Reference: Lecture Notes Calculus 1 ( May22,2022 ) Definition8.1 page 144

This applet visualizes pointconvergence using the sequence of functions

The blue solid curve is the current function $f_n$. Faint dashed blue curves show earlier functions $x^k$ (only for $k<n$) to make the trend visible. The pointwise limit is drawn in red: the line $y=0$ (the “zero function”), together with the special point $(1,1)$, since $1^n=1$ for all $n$.

Two sliders are used: n selects the current index $n$ in $f_n(x)=x^n$, and ε selects the tolerance $\varepsilon>0$. A draggable point x chooses the evaluation point $x\in[0,1]$. The applet draws the horizontal $\varepsilon$-tube $f(x)\pm\varepsilon$ and highlights the point $(x,f_n(x))$. This point turns green exactly when

The displayed value $N_{x,\varepsilon}$ is the first index after which the inequality holds at the chosen $x$ and $\varepsilon$.

const ok = Math.abs(fnx - fx) < eps;

First, increase $n$ and observe that for any fixed $x < 1$, the values $x^n$ drop toward $0$.
Hence, the blue curve approaches the red zero function on $[0,1)$.

The behaviour at $x = 1$ is different:

which is why the limit includes the point $(1,1)$.