From 773e8508730f5d92cc9b8bce5339669f65ddd064 Mon Sep 17 00:00:00 2001 From: arc Date: Thu, 6 Mar 2025 09:14:22 -0700 Subject: [PATCH] vault backup: 2025-03-06 09:14:22 --- education/math/MATH1210 (calc 1)/Limits.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/education/math/MATH1210 (calc 1)/Limits.md b/education/math/MATH1210 (calc 1)/Limits.md index f49ee84..31cd4f7 100644 --- a/education/math/MATH1210 (calc 1)/Limits.md +++ b/education/math/MATH1210 (calc 1)/Limits.md @@ -82,6 +82,7 @@ If you have a limit of the indeterminate form $\dfrac{0}{0}$, the limit can be f $$ \lim_{x \to 2} \dfrac{x-2}{x^2-4} = \lim_{x \to 2} \dfrac{1}{2x}$$ L'Hospital's Rule can also be used when both the numerator and denominator approach some form of infinity. -$$ $$ +$$ \lim_{x \to \infty} \dfrac{x^2-2}{3x^2-4} = \lim_{x \to \infty} \dfrac{2x}{6x}$$ +The above problem can be solved more easily *without* L'Hospital's rule, the leading coefficients are 1/3, so the limit as $x$ approaches $\infty$ is 1/3. L'Hospital's rule **cannot** be used in any other circumstance. \ No newline at end of file