use plotly; fix bitrot

quarto/limits/limits.qmd

@@ -9,6 +9,7 @@ This section uses the following add-on packages:
 ```{julia}
 using CalculusWithJulia
 using Plots
+plotly()
 using Richardson  # for extrapolation
 using SymPy       # for symbolic limits
 ```
@@ -28,7 +29,7 @@ There wasn't a ready-made formula for the area of this shape, as was known for a
 #| echo: false
 #| cache: true
 ###{{{archimedes_parabola}}}
-
+gr()
 f(x) = x^2
 colors = [:black, :blue, :orange, :red, :green, :orange, :purple]
 
@@ -76,6 +77,7 @@ The first triangle has area $1/2$, the second has area $1/8$, then $2$ have area
 With some algebra, the total area then should be $1/2 \cdot (1 + (1/4) + (1/4)^2 + \cdots) = 2/3$.
 """
 
+plotly()
 ImageFile(imgfile, caption)
 ```
 
@@ -102,7 +104,7 @@ Next, we illustrate how Archimedes approximated $\pi$ – the ratio of the circu
 #| hold: true
 #| echo: false
 ## Archimedes approximation for pi
-
+gr()
 blue, green, purple, red = :royalblue, :forestgreen, :mediumorchid3, :brown3
 
 
@@ -161,7 +163,7 @@ end
 caption = L"""
 The ratio of the circumference of a circle to its diameter, $\pi$, can be approximated from above and below by computing the perimeters of the inscribed $n$-gons. Archimedes computed the perimeters for $n$ being $12$, $24$, $48$, and $96$ to determine that $3~1/7 \leq \pi \leq 3~10/71$.
 """
-
+plotly()
 ImageFile(p, caption)
 ```
 
@@ -217,6 +219,7 @@ The above bound comes from this figure, for small $x > 0$:
 ```{julia}
 #| hold: true
 #| echo: false
+gr()
 p = plot(x -> sqrt(1 - x^2), 0, 1, legend=false, aspect_ratio=:equal,
      linewidth=3, color=:black)
 θ = π/6
@@ -237,6 +240,7 @@ imgfile = tempname() * ".png"
 savefig(p, imgfile)
 caption = "Triangle ``ABD` has less area than the shaded wedge, which has less area than triangle ``ACD``. Their respective areas are ``(1/2)\\sin(\\theta)``, ``(1/2)\\theta``, and ``(1/2)\\tan(\\theta)``. The inequality used to show ``\\sin(x)/x`` is bounded below by ``\\cos(x)`` and above by ``1`` comes from a division by ``(1/2) \\sin(x)`` and taking reciprocals.
 "
+plotly()
 ImageFile(imgfile, caption)
 ```
 
@@ -1022,7 +1026,7 @@ Then the limit of $f$ must also  be $L$.
 ```{julia}
 #| hold: true
 #| echo: false
-
+gr()
 function squeeze_example(x)
 	x₀ = 0.5
 	plot(cos, 0, x₀, label="cos")
@@ -1043,6 +1047,7 @@ gif(anim, imgfile, fps = 1)
 caption = """
 As ``x`` goes to ``0``, the values of ``\\sin(x)/x`` are squeezed between ``\\cos(x)`` and ``1`` which both converge to ``1``.
 """
+plotly()
 ImageFile(imgfile, caption)
 ```
 
@@ -1089,6 +1094,7 @@ In this case, $\delta$ is easy to guess, as the function is linear and has slope
 #| echo: false
 #| cache: true
 ## {{{ limit_e_d }}}
+gr()
 function make_limit_e_d(n)
     f(x) = x^3
 
@@ -1135,7 +1141,7 @@ does not leave the top or bottom of the box (except possibly at the
 edges). In this example $\delta^3=\epsilon$.
 
 """
-
+plotly()
 ImageFile(imgfile, caption)
 ```
 
@@ -1247,6 +1253,7 @@ Select the graph for which there is no limit at $a$.
 ```{julia}
 #| hold: true
 #| echo: false
+gr()
 let
     p1 = plot(;axis=nothing, legend=false)
     title!(p1, "(a)")
@@ -1281,6 +1288,11 @@ let
 end
 ```
 
+```{julia}
+#| echo: false
+plotly();
+```
+
 ###### Question