small edits

quarto/limits/limits.qmd

@@ -1532,6 +1532,43 @@ numericq(val)
 ###### Question
 
 
+Compute the limit
+
+
+$$
+\lim_{x \rightarrow 0} \frac{x\sin(\sin(x)) - \sin^2(x)}{x^6}.
+$$
+
+```{julia}
+#| hold: true
+#| echo: false
+f(x) =  (x * sin(sin(x))- sin(x)^2)/x^6
+val = N(limit(f(x), x => 0))
+numericq(val)
+```
+
+
+###### Question
+
+
+Compute the limit
+
+
+$$
+\lim_{x \rightarrow 0} \frac{\tan(x) - 24 * \tan(x/2)}{4  \sin(x) - 5  x}.
+$$
+
+```{julia}
+#| hold: true
+#| echo: false
+f(x) = (tan(x) - 24 * tan(x/2)) / (4 * sin(x) - 5 * x)
+val = N(limit(f(x), x => 0))
+numericq(val)
+```
+
+###### Question
+
+
 Some limits involve parameters. For example, suppose we define `ex` as follows:
 
 
@@ -1616,6 +1653,47 @@ Should `SymPy` have needed an assumption like
 yesnoq("yes")
 ```
 
+###### Question
+
+The limit
+
+$$
+L= \lim_{x \rightarrow 0} \left(\frac{(a^x - x \log(a)}{b^x - x\log(b)}\right)^{1/x^2}
+$$
+
+For $a=3$ and $b=2$
+
+Can be computed symbolically *two* different ways:
+
+```{julia}
+@syms x
+a, b = 3, 2
+f(x) = ((a^x - x*log(a))/(b^x - x*log(b)))^(1/x^2)
+limit(f(x), x=>0)
+```
+
+*or*
+
+```{julia}
+@syms x a b
+f(x) = ((a^x - x*log(a))/(b^x - x*log(b)))^(1/x^2)
+L = limit(f(x), x=>0)
+L(a => 3, b=>2)
+```
+
+Which is correct?
+
+```{julia}
+#| echo: false
+choices = ["The first one", "The second one"]
+explanation = """
+The first one is incorrect, as `log(3)` is evaluated numerically, and not symbolically. The difference between a floating point approximation and the symbolic expression is enough to make the first limit infinite, despite the actual limit being finite.
+"""
+buttonq(choices, 2; explanation=explanation)
+```
+
+
+
 ###### Question: The squeeze theorem