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quarto/derivatives/lhospitals_rule.qmd

@@ -307,24 +307,24 @@ L'Hospital's rule generalizes to other indeterminate forms, in particular the in
 The value $c$ in the limit can also be infinite. Consider this case with $c=\infty$:
 
 
-$$
+
 \begin{align*}
 \lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} &=
 \lim_{x \rightarrow 0} \frac{f(1/x)}{g(1/x)}
 \end{align*}
-$$
+
 
 L'Hospital's limit applies as $x \rightarrow 0$, so we differentiate to get:
 
 
-$$
+
 \begin{align*}
 \lim_{x \rightarrow 0} \frac{[f(1/x)]'}{[g(1/x)]'}
 &= \lim_{x \rightarrow 0} \frac{f'(1/x)\cdot(-1/x^2)}{g'(1/x)\cdot(-1/x^2)}\\
 &= \lim_{x \rightarrow 0} \frac{f'(1/x)}{g'(1/x)}\\
 &= \lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)},
 \end{align*}
-$$
+
 
 *assuming* the latter limit exists, L'Hospital's rule assures the equality
 
@@ -414,12 +414,12 @@ Be just saw that $\lim_{x \rightarrow 0+}\log(x)/(1/x) = 0$. So by the rules for
 A limit $\lim_{x \rightarrow c} f(x) - g(x)$ of indeterminate form $\infty - \infty$ can be reexpressed to be of the from $0/0$ through the transformation:
 
 
-$$
+
 \begin{align*}
 f(x) - g(x) &= f(x)g(x) \cdot (\frac{1}{g(x)} - \frac{1}{f(x)}) \\
 &= \frac{\frac{1}{g(x)} - \frac{1}{f(x)}}{\frac{1}{f(x)g(x)}}.
 \end{align*}
-$$
+
 
 Applying this to