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jverzani
2025-07-23 08:05:43 -04:00
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quarto/limits/limits.qmd
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@@ -36,27 +36,38 @@ colors = [:black, :blue, :orange, :red, :green, :orange, :purple]
## Area of parabola
function make_triangle_graph(n)
title = "Area of parabolic cup ..."
n==1 && (title = "\${Area = }1/2\$")
n==2 && (title = "\${Area = previous }+ 1/8\$")
n==3 && (title = "\${Area = previous }+ 2\\cdot(1/8)^2\$")
n==4 && (title = "\${Area = previous }+ 4\\cdot(1/8)^3\$")
n==5 && (title = "\${Area = previous }+ 8\\cdot(1/8)^4\$")
n==6 && (title = "\${Area = previous }+ 16\\cdot(1/8)^5\$")
n==7 && (title = "\${Area = previous }+ 32\\cdot(1/8)^6\$")
n==1 && (title = L"Area $= 1/2$")
n==2 && (title = L"Area $=$ previous $+\; \frac{1}{8}$")
n==3 && (title = L"Area $=$ previous $+\; 2\cdot\frac{1}{8^2}$")
n==4 && (title = L"Area $=$ previous $+\; 4\cdot\frac{1}{8^3}$")
n==5 && (title = L"Area $=$ previous $+\; 8\cdot\frac{1}{8^4}$")
n==6 && (title = L"Area $=$ previous $+\; 16\cdot\frac{1}{8^5}$")
n==7 && (title = L"Area $=$ previous $+\; 32\cdot\frac{1}{8^6}$")
plt = plot(f, 0, 1, legend=false, size = fig_size, linewidth=2)
annotate!(plt, [(0.05, 0.9, text(title,:left))]) # if in title, it grows funny with gr
n >= 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1], color=colors[1], linetype=:polygon, fill=colors[1], alpha=.2)
n == 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1], color=colors[1], linewidth=2)
plt = plot(f, 0, 1;
legend=false,
size = fig_size,
linewidth=2)
annotate!(plt, [
(0.05, 0.9, text(title,:left))
]) # if in title, it grows funny with gr
n >= 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1];
color=colors[1], linetype=:polygon,
fill=colors[1], alpha=.2)
n == 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1];
color=colors[1], linewidth=2)
for k in 2:n
xs = range(0,stop=1, length=1+2^(k-1))
ys = map(f, xs)
k < n && plot!(plt, xs, ys, linetype=:polygon, fill=:black, alpha=.2)
xs = range(0, stop=1, length=1+2^(k-1))
ys = f.(xs)
k < n && plot!(plt, xs, ys;
linetype=:polygon, fill=:black, alpha=.2)
if k == n
plot!(plt, xs, ys, color=colors[k], linetype=:polygon, fill=:black, alpha=.2)
plot!(plt, xs, ys, color=:black, linewidth=2)
plot!(plt, xs, ys;
color=colors[k], linetype=:polygon, fill=:black, alpha=.2)
plot!(plt, xs, ys;
color=:black, linewidth=2)
end
end
plt
@@ -1041,18 +1052,34 @@ $$
Why? We can express the function $e^{\csc(x)}/e^{\cot(x)}$ as the above function plus the polynomial $1 + x/2 + x^2/8$. The above is then the sum of two functions whose limits exist and are finite, hence, we can conclude that $M = 0 + 1$.
### The [squeeze](http://en.wikipedia.org/wiki/Squeeze_theorem) theorem
### The squeeze theorem
Sometimes limits can be found by bounding more complicated functions by easier functions.
::: {.callout-note icon=false}
## The [squeeze theorem](http://en.wikipedia.org/wiki/Squeeze_theorem)
Fix $c$ in $I=(a,b)$. Suppose for all $x$ in $I$, except possibly $c$, there are two functions $l$ and $u$, satisfying:
We note one more limit law. Suppose we wish to compute $\lim_{x \rightarrow c}f(x)$ and we have two other functions, $l$ and $u$, satisfying:
* $l(x) \leq f(x) \leq u(x)$.
* These limits exist and are equal:
$$
L = \lim_{x \rightarrow c} l(x) = \lim_{x \rightarrow c} u(x).
$$
* for all $x$ near $c$ (possibly not including $c$) $l(x) \leq f(x) \leq u(x)$.
* These limits exist and are equal: $L = \lim_{x \rightarrow c} l(x) = \lim_{x \rightarrow c} u(x)$.
Then
$$
\lim_{x\rightarrow c} f(x) = L.
$$
Then the limit of $f$ must also be $L$.
:::
The figure shows a usage of the squeeze theorem to show $\sin(x)/x \rightarrow 1$ as $\cos(x) \leq \sin(x)x \leq 1$ for $x$ close to $0$.
```{julia}
#| hold: true
@@ -1088,8 +1115,71 @@ ImageFile(imgfile, caption)
The formal definition of a limit involves clarifying what it means for $f(x)$ to be "close to $L$" when $x$ is "close to $c$". These are quantified by the inequalities $0 < \lvert x-c\rvert < \delta$ and the $\lvert f(x) - L\rvert < \epsilon$. The second does not have the restriction that it is greater than $0$, as indeed $f(x)$ can equal $L$. The order is important: it says for any idea of close for $f(x)$ to $L$, an idea of close must be found for $x$ to $c$.
The key is identifying a value for $\delta$ for a given value of $\epsilon$.
::: {#fig-limit-e-d}
```{julia}
#| echo: false
plt = let
gr()
f(x) = (x+1)^2 -1
a, b = -1/4, 2
c = 1
L = f(c)
δ = 0.2
ϵ = 3sqrt(δ)
plot(; empty_style...)#, aspect_ratio=:equal)
plot!(f, a, b; line=(:black, 2))
plot!([a,b],[0,0]; axis_style...)
plot!([0,0], [f(a), f(2)]; axis_style...)
plot!([c, c, 0], [0, f(c), f(c)]; line=(:black, 1, :dash))
plot!([c-δ, c-δ, 0], [0, f(c-δ), f(c-δ)]; line=(:black, 1, :dashdot))
plot!([c+δ, c+δ, 0], [0, f(c+δ), f(c+δ)]; line=(:black, 1, :dashdot))
S = Shape([0,b,b,0],[L-ϵ,L-ϵ,L+ϵ,L+ϵ])
plot!(S; fill=(:lightblue, 0.25), line=(nothing,))
domain_color=:red
range_color=:blue
S = Shape([c-δ, c+δ, c+δ, c-δ], 0.05*[-1,-1,1,1])
plot!(S, fill=(domain_color,0.4), line=nothing)
m,M = f(c-δ), f(c+δ)
T = Shape(0.015 * [-1,1,1,-1], [m,m,M,M])
plot!(T, fill=(range_color, 0.4), line=nothing)
C = Plots.scale(Shape(:circle), 0.02, 0.1)
plot!(Plots.translate(C, c, L), fill=(:white,1,0), line=(:black, 1))
plot!(Plots.translate(C, c, 0), fill=(:white,1,0), line=(domain_color, 1))
plot!(Plots.translate(C, 0, L), fill=(:white,1,0), line=(range_color, 1))
annotate!([
(c, 0, text(L"c", :top)),
(c-δ, 0, text(L"c - \delta", 10, :top)),
(c+δ, 0, text(L"c + \delta", 10, :top)),
(0, L, text(L"L", :right)),
(0, L+ϵ, text(L"L + \epsilon", 10, :right)),
(0, L-ϵ, text(L"L - \epsilon", 10, :right)),
])
end
plt
```
```{julia}
#| echo: false
plotly()
nothing
```
Figure illustrating requirements of $\epsilon-\delta$ definition of the limit. The image (shaded red on $y$ axis) of the $x$ within $\delta$ of $c$ (except for $c$ and shaded blue on the $x$ axis) must stay within the bounds of $L-\epsilon$ and $L+ \epsilon$, where $\delta$ may be chosen based on $\epsilon$ but needs to be chosen for every positive $\epsilon$, not just a fixed one as in this figure.
:::
A simple case is the linear case. Consider the function $f(x) = 3x + 2$. Verify that the limit at $c=1$ is $5$.
@@ -1117,64 +1207,78 @@ These lines produce a random $\epsilon$, the resulting $\delta$, and then verify
(The random numbers are technically in $[0,1)$, so in theory `epsilon` could be `0`. So the above approach would be more solid if some guard, such as `epsilon = max(eps(), rand())`, was used. As the formal definition is the domain of paper-and-pencil, we don't fuss.)
In this case, $\delta$ is easy to guess, as the function is linear and has slope $3$. This basically says the $y$ scale is 3 times the $x$ scale. For non-linear functions, finding $\delta$ for a given $\epsilon$ can be a challenge. For the function $f(x) = x^3$, illustrated below, a value of $\delta=\epsilon^{1/3}$ is used for $c=0$:
In this case, $\delta$ is easy to guess, as the function is linear and has slope $3$. This basically says the $y$ scale is 3 times the $x$ scale. For non-linear functions, finding $\delta$ for a given $\epsilon$ can be more of a challenge.
```{julia}
#| hold: true
#| echo: false
#| cache: true
## {{{ limit_e_d }}}
gr()
function make_limit_e_d(n)
f(x) = x^3
##### Example
xs = range(-.9, stop=.9, length=50)
ys = map(f, xs)
We show using the definition that for any fixed $a$ and $n$:
$$
\lim_{x \rightarrow a} x^n = a^n.
$$
This proof uses a bound based on properties of the absolute value.
plt = plot(f, -.9, .9, legend=false, size=fig_size)
if n == 0
nothing
else
k = div(n+1,2)
epsilon = 1/2^k
delta = cbrt(epsilon)
if isodd(n)
plot!(plt, xs, 0*xs .+ epsilon, color=:orange)
plot!(plt, xs, 0*xs .- epsilon, color=:orange)
else
plot!(delta * [-1, 1], epsilon * [ 1, 1], color=:orange)
plot!(delta * [ 1, -1], epsilon * [-1,-1], color=:orange)
plot!(delta * [-1, -1], epsilon * [-1, 1], color=:red)
plot!(delta * [ 1, 1], epsilon * [-1, 1], color=:red)
end
end
plt
end
We look at $f(x) - L = x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + \cdots + x^1a^{n-1} + a^n)$.
Taking absolute values gives an inequality by the triangle inequality:
n = 11
anim = @animate for i=1:n
make_limit_e_d(i-1)
end
$$
\lvert x^n - a^n\rvert \leq \lvert x-a\rvert\cdot
\left(
\lvert x^{n-1}\rvert +
\lvert x^{n-2}\rvert\lvert a\rvert +
\cdots +
\lvert x^1\rvert\lvert a^{n-1}\rvert +
\lvert a^n\rvert
\right).
$$
imgfile = tempname() * ".gif"
gif(anim, imgfile, fps = 1)
Now, for a given $\epsilon>0$ we seek a $\delta>0$ satisfying the properties of the limit definition for $f(x) = x^n$ and $L=a^n$. For now, assume $\delta < 1$. Then we can assume $\lvert x-a\rvert < \delta$ and
$$
\lvert x\rvert = \lvert x - a + a\rvert \leq \lvert x-a\rvert + \lvert a\rvert < 1 + \lvert a\rvert
$$
caption = L"""
This says then
Demonstration of $\epsilon$-$\delta$ proof of $\lim_{x \rightarrow 0}
x^3 = 0$. For any $\epsilon>0$ (the orange lines) there exists a
$\delta>0$ (the red lines of the box) for which the function $f(x)$
does not leave the top or bottom of the box (except possibly at the
edges). In this example $\delta^3=\epsilon$.
$$
\begin{align*}
\lvert x^n - a^n\rvert
&\leq
\lvert x-a\rvert\cdot \left(
\lvert x\rvert^{n-1} +
\lvert x\rvert^{n-2}\lvert a\rvert +
\cdots +
\lvert x\rvert^1\lvert a^{n-1}\rvert +
\lvert a^n\rvert
\right)\\
%%
&\leq \lvert x - a\rvert
\cdot \left(
(\lvert a\rvert+1)^{n-1} +
(\lvert a\rvert+1)^{n-2}\lvert a\rvert
+ \cdots +
(\lvert a\rvert+1)^1 \lvert a^{n-1}\rvert +
\lvert a^n \rvert
\right)\\
&\leq \lvert x-a\rvert \cdot C,
\end{align*}
$$
where $C$ is just some constant not depending on $x$, just $a$ and $n$.
Now if $\delta < 1$ and $\delta < \epsilon/C$ and if
$0 < \lvert x - a \rvert < \delta$ then
$$
\lvert f(x) - L \rvert =
\lvert x^n - a^n\rvert \leq \lvert x-a\rvert \cdot C \leq \delta\cdot C < \frac{\epsilon}{C} \cdot C = \epsilon.
$$
With this result, the rules of limits can immediately extend this to any polynomial, $p(x),$ it follows that $\lim_{x \rightarrow c} p(x) = p(a)$. (Because $c_n x^n \rightarrow c_n a^n$ and the sum of two functions with a limit has the limit of the sums.) Based on this, we will say later that any polynomial is *continuous* for all $x$.
"""
plotly()
ImageFile(imgfile, caption)
```
## Questions
@@ -1290,26 +1394,26 @@ let
title!(p1, "(a)")
plot!(p1, x -> x^2, 0, 2, color=:black)
plot!(p1, zero, linestyle=:dash)
annotate!(p1,[(1,0,"a")])
annotate!(p1,[(1,0,text(L"a",:top))])
p2 = plot(;axis=nothing, legend=false)
title!(p2, "(b)")
plot!(p2, x -> 1/(1-x), 0, .95, color=:black)
plot!(p2, x-> -1/(1-x), 1.05, 2, color=:black)
plot!(p2, zero, linestyle=:dash)
annotate!(p2,[(1,0,"a")])
annotate!(p2,[(1,0,text(L"a",:top))])
p3 = plot(;axis=nothing, legend=false)
title!(p3, "(c)")
plot!(p3, sinpi, 0, 2, color=:black)
plot!(p3, zero, linestyle=:dash)
annotate!(p3,[(1,0,"a")])
annotate!(p3,[(1,0,text("a",:top))])
p4 = plot(;axis=nothing, legend=false)
title!(p4, "(d)")
plot!(p4, x -> x^x, 0, 2, color=:black)
plot!(p4, zero, linestyle=:dash)
annotate!(p4,[(1,0,"a")])
annotate!(p4,[(1,0,text(L"a",:top))])
l = @layout[a b; c d]
p = plot(p1, p2, p3, p4, layout=l)
@@ -1520,8 +1624,8 @@ Take
$$
f(x) = \begin{cases}
0 & x \neq 0\\
1 & x = 0
0 &~ x \neq 0\\
1 &~ x = 0
\end{cases}
$$