em dash; sentence case

quarto/limits/intermediate_value_theorem.qmd

@@ -315,7 +315,7 @@ find_zero(q, (5, 10))
 ::: {.callout-note}
 ### Between need not be near
 
-Later, we will see more efficient algorithms to find a zero *near* a given guess. The bisection method finds a zero *between* two values of a bracketing interval. This interval need not be small. Indeed in many cases it can be infinite. For this particular problem, any interval like `(2,N)` will work as long as `N` is bigger than the zero and small enough that `q(N)` is finite *or* infinite *but* not `NaN`. (Basically, `q` must evaluate to a number with a sign. Here, the value of `q(Inf)` is `NaN` as it evaluates to the indeterminate `Inf - Inf`. But `q` is still not `NaN` for quite large numbers, such as `1e77`, as `x^4` can as big as `1e308` -- technically `floatmax(Float64)` -- and be finite.)
+Later, we will see more efficient algorithms to find a zero *near* a given guess. The bisection method finds a zero *between* two values of a bracketing interval. This interval need not be small. Indeed in many cases it can be infinite. For this particular problem, any interval like `(2,N)` will work as long as `N` is bigger than the zero and small enough that `q(N)` is finite *or* infinite *but* not `NaN`. (Basically, `q` must evaluate to a number with a sign. Here, the value of `q(Inf)` is `NaN` as it evaluates to the indeterminate `Inf - Inf`. But `q` is still not `NaN` for quite large numbers, such as `1e77`, as `x^4` can as big as `1e308`---technically `floatmax(Float64)`---and be finite.)
 
 :::
 
@@ -840,7 +840,7 @@ plotly()
 nothing
 ```
 
-Figure illustrating absolute and relative minima for a function $f(x)$ over $I=[a,b]$. The leftmost point has a $y$ value, $f(a)$, which is an absolute maximum of $f(x)$ over $I$. The three points highlighted between $a$ and $b$ are all relative extrema. The first one is *also* the absolute minimum over $I$. The endpoint is not considered a relative maximum for technical reasons --- there is no interval around $b$, it being on the boundary of $I$.
+Figure illustrating absolute and relative minima for a function $f(x)$ over $I=[a,b]$. The leftmost point has a $y$ value, $f(a)$, which is an absolute maximum of $f(x)$ over $I$. The three points highlighted between $a$ and $b$ are all relative extrema. The first one is *also* the absolute minimum over $I$. The endpoint is not considered a relative maximum for technical reasons---there is no interval around $b$, it being on the boundary of $I$.
 
 :::
 

quarto/limits/limits.qmd

@@ -689,7 +689,7 @@ c = 15/11
 lim(h, c; n = 16)
 ```
 
-(Though the graph and table do hint at something a bit odd -- the graph shows a blip, the table doesn't show values in the second column going towards a specific value.)
+(Though the graph and table do hint at something a bit odd---the graph shows a blip, the table doesn't show values in the second column going towards a specific value.)
 
 
 However the limit in this case is $-\infty$ (or DNE), as there is an aysmptote at $c=15/11$. The problem is the asymptote due to the logarithm is extremely narrow and happens between floating point values to the left and right of $15/11$.