many edits

quarto/derivatives/lhospitals_rule.qmd

@@ -229,7 +229,7 @@ Of course, we could have just relied on `limit`, which knows about L'Hospital's
 
 
 ```{julia}
-limit(f(x)/g(x), x, a)
+limit(f(x)/g(x), x => a)
 ```
 
 ## Idea behind L'Hospital's rule
@@ -272,13 +272,13 @@ function lhopitals_picture_graph(n)
     plt
 end
 
-caption = L"""
-
-Geometric interpretation of ``L=\lim_{x \rightarrow 0} x^2 / (\sqrt{1 +
-x} - 1 - x^2)``. At ``0`` this limit is indeterminate of the form
+caption = raw"""
+Geometric interpretation of
+``L=\lim_{x \rightarrow 0} x^2 / (\sqrt{1 + x} - 1 - x^2)``.
+At ``0`` this limit is indeterminate of the form
 ``0/0``. The value for a fixed ``x`` can be seen as the slope of a secant
-line of a parametric plot of the two functions, plotted as ``(g,
-f)``. In this figure, the limiting "tangent" line has ``0`` slope,
+line of a parametric plot of the two functions, plotted as
+``(g, f)``. In this figure, the limiting "tangent" line has ``0`` slope,
 corresponding to the limit ``L``. In general, L'Hospital's rule is
 nothing more than a statement about slopes of tangent lines.
 
@@ -437,38 +437,38 @@ $$
 \lim_{x \rightarrow 1} \frac{x\log(x)-(x-1)}{(x-1)\log(x)}
 $$
 
-In `SymPy` we have:
+In `SymPy` we have (using italic variable names to avoid a problem with the earlier use of `f` as a function):
 
 
 ```{julia}
-𝒇 = x*log(x) - (x-1)
-𝒈 = (x-1)*log(x)
-𝒇(1), 𝒈(1)
+𝑓 = x * log(x) - (x-1)
+𝑔 = (x-1) * log(x)
+𝑓(1), 𝑔(1)
 ```
 
 L'Hospital's rule applies to the form $0/0$, so we try:
 
 
 ```{julia}
-𝒇′ = diff(𝒇, x)
-𝒈′ = diff(𝒈, x)
-𝒇′(1), 𝒈′(1)
+𝑓′ = diff(𝑓, x)
+𝑔′ = diff(𝑔, x)
+𝑓′(1), 𝑔′(1)
 ```
 
 Again, we get the indeterminate form $0/0$, so we try again with second derivatives:
 
 
 ```{julia}
-𝒇′′ = diff(𝒇, x, x)
-𝒈′′ = diff(𝒈, x, x)
-𝒇′′(1), 𝒈′′(1)
+𝑓′′ = diff(𝑓, x, x)
+𝑔′′ = diff(𝑔, x, x)
+𝑓′′(1), 𝑔′′(1)
 ```
 
 From this we see the limit is $1/2$, as could have been done directly:
 
 
 ```{julia}
-limit(𝒇/𝒈, x=>1)
+limit(𝑓/𝑔, x=>1)
 ```
 
 ## The assumptions are necessary
@@ -524,7 +524,7 @@ This ratio has no limit, as it oscillates, as confirmed by `SymPy`:
 
 
 ```{julia}
-limit(u(x)/v(x), x=> oo)
+limit(u(x)/v(x), x => oo)
 ```
 
 ## Questions
@@ -645,8 +645,8 @@ What is $L$?
 #| hold: true
 #| echo: false
 f(x) = (4x - sin(x))/x
-L = float(N(limit(f, 0)))
-numericq(L)
+L = limit(f(x), x=>0)
+numericq(float(L))
 ```
 
 ###### Question
@@ -666,8 +666,8 @@ What is $L$?
 #| hold: true
 #| echo: false
 f(x) = (sqrt(1+x) - 1)/x
-L = float(N(limit(f, 0)))
-numericq(L)
+L = limit(f(x), x => 0)
+numericq(float(L))
 ```
 
 ###### Question
@@ -687,8 +687,8 @@ What is $L$?
 #| hold: true
 #| echo: false
 f(x) = (x - sin(x))/x^3
-L = float(N(limit(f, 0)))
-numericq(L)
+L = limit(f(x), x=>0)
+numericq(float(L))
 ```
 
 ###### Question
@@ -708,8 +708,8 @@ What is $L$?
 #| hold: true
 #| echo: false
 f(x) = (1 - x^2/2 - cos(x))/x^3
-L = float(N(limit(f, 0)))
-numericq(L)
+L = limit(f(x), x=>0)
+numericq(float(L))
 ```
 
 ###### Question
@@ -729,8 +729,8 @@ What is $L$?
 #| hold: true
 #| echo: false
 f(x) = log(log(x))/log(x)
-L = N(limit(f(x), x=> oo))
-numericq(L)
+L = limit(f(x), x=> oo)
+numericq(float(L))
 ```
 
 ###### Question
@@ -750,8 +750,8 @@ What is $L$?
 #| hold: true
 #| echo: false
 f(x) = 1/x - 1/sin(x)
-L = float(N(limit(f, 0)))
-numericq(L)
+L = limit(f(x), x => 0)
+numericq(float(L))
 ```
 
 ###### Question
@@ -770,8 +770,8 @@ What is $L$?
 ```{julia}
 #| hold: true
 #| echo: false
-L = float(N(limit(log(x)/x, x=>oo)))
-numericq(L)
+L = limit(log(x)/x, x=>oo)
+numericq(float(L))
 ```
 
 ###### Question