vault backup: 2025-03-06 09:44:22

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arc 2025-03-06 09:44:22 -07:00
parent cfda79dc6a
commit da92f848f3
2 changed files with 29 additions and 2 deletions

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@ -0,0 +1,27 @@
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@ -88,7 +88,7 @@ The above problem can be solved more easily *without* L'Hospital's rule, the lea
L'Hospital's rule **cannot** be used in any other circumstance.
## Examples
1. $\lim_{x \ to 0} \dfrac{7^x - 5^x}{2x}$
1. $\lim_{x \to 0} \dfrac{7^x - 5^x}{2x}$
2. $= \lim_{x \ to 0}\dfrac{7^x \ln(7) -5^x(\ln(5)}{2}$
3. $= \dfrac{\ln(7) - \ln(5)}{2}$
# Indeterminate form $(0 * \infty)$
@ -96,4 +96,4 @@ If the $\lim_{x \to a}f(x) = 0$ and $\lim_{x\to a} g(x) = \infty$ then $\lim_{x
To evaluate an indeterminate product ($0 * \infty$), use algebra to convert the product to an equivalent quotient and then use L'Hopsital's Rule.
$$ \lim_{x \to 0^+} x\ln(x) = \lim_{x \to 0^+}\dfrac{\ln x}{\dfrac{1}{x}} = \\lim_{x \to 0^+} \dfrac{1/x}{-1/(x^2)} = \lim_{x \to 0^+} -x = 0 $$
$$ \lim_{x \to 0^+} x\ln(x) = \lim_{x \to 0^+}\dfrac{\ln x}{\dfrac{1}{x}} = \lim_{x \to 0^+} \dfrac{1/x}{-1/(x^2)} = \lim_{x \to 0^+} -x = 0 $$