> [!summary] L'Hôpital's rule is a way of solving limits of a specific indeterminate form, $\frac{0}{0}$ or $\frac{\infty}{\infty}$ > **Key Equations:** $\displaystyle \lim_{ x \to a } \frac{f (x)}{g (x)} = \displaystyle \lim_{ x \to a } \frac{f' (x)}{g' (x)} = \frac{f' (a)}{g (a)} \quad \text{if } g' (a)\neq 0$ >[!info]+ Read Time **⏱ 3 mins** # Definition L'Hôpital's Rule is a way of solving limits that are in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, an [[Indeterminate Forms|indeterminate form]]. This means a limit like $\displaystyle \lim_{ x \to a } \frac{f(x)}{g(x)}=\frac{0}{0} \space\text{or} =\frac{\infty}{\infty}$. This rule states that given that $\frac{f(a)}{g(a)}$ are [[Indeterminate Forms|undefined]], then the limit is must be equal to the [[Derivative|derivative]] of the functions ($f(x),g(x)$), defined below. $ \displaystyle \lim_{ x \to a } \frac{f(x)}{g(x)} = \displaystyle \lim_{ x \to a } \frac{f'(x)}{g'(x)} = \frac{f'(a)}{g(a)} \quad \text{if } g'(a)\neq 0 $ > [!note] This rule [[Differentiable|differentiates]] the two functions separately (no quotient). As well, this rule doesn't guarantee that the [[Limits|limits]] of the [[Differentiable|differentiable]] are defined ## Proof > [!warning] Assumptions To show a simplified proof for L'Hospital's rule through [[Direct Proof|direct proof]] assume the following: > - A [[Derivative|derivative]] is defined as $f'(x)=\displaystyle \lim_{ x\to a } \frac{f (x) - f (a)}{x-a}$ > - The [[Analytical Limits Rules|quotient limit rule]] states $\displaystyle \lim_{ n \to a } \frac{f(x)}{g(x)}=\frac{\displaystyle \lim_{ n \to a } f (x)}{\displaystyle \lim_{ n \to a } g (x)}$ > [!note]+ This proof is only for the case that $\displaystyle \lim_{ x \to a } \frac{f(x)}{g(x)}=\frac{0}{0}$, but is similar approach is done for the $\displaystyle \lim_{ x \to a } \frac{f(x)}{g(x)}=\frac{\infty}{\infty}$, which is not proven here. This proof avoids using the [[Epsilon-Delta Definition of a Limit|epsilon-delta definition of a limit]], for resources on that refer to [[L'Hôpital's Rule#Resources|resources]]. Prove that $\displaystyle \lim_{ x \to a } \frac{f(x)}{g(x)} = \displaystyle \lim_{ x \to a } \frac{f'(x)}{g'(x)} =\frac{f'(a)}{g'(a)}$ if $f(a)=g(a)=0$ $ \begin{array}{c} \text{Given that} \\ \displaystyle \lim_{ x \to a } \frac{f(x)}{g(x)} \\ \\ \text{This is the same as saying} \\ \displaystyle \lim_{ x \to a } \frac{f(x)-0}{g(x)-0} \\ \\ \text{Since $f(a)=g(a)=0$, then} \\ \displaystyle \lim_{ x \to a } \frac{f(x)-0}{g(x)-0} = \displaystyle \lim_{ x \to a } \frac{f(x)-f(a)}{g(x)-g(a)} \\ \\ \text{Then you can divide both top and bottom by $x-a$, since it doesnt change the equation} \\ \displaystyle \lim_{ x \to a } \frac{f(x)-f(a)}{g(x)-g(a)} = \displaystyle \lim_{ x \to a } \frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}} = \frac{ \displaystyle \lim_{ x \to a } \frac{f(x)-f(a)}{x-a}}{ \displaystyle \lim_{ x \to a } \frac{g(x)-g(a)}{x-a}} = \frac{ \displaystyle \lim_{ x \to a } f'(x) }{\displaystyle \lim_{ x \to a } g'(x) }= \frac{f'(a)}{g'(a)} \end{array} $ # Resources <iframe width="560" height="315" src="https://www.youtube.com/embed/AiQUe8M8dj8?si=UT8bKqyiZ7o48-Ee" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe> <iframe width="560" height="315" src="https://www.youtube.com/embed/tL13JxmyRLw?si=ygUl5jz3Hb3-DUFe" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe> Uses epsilon-delta proofing --- > 🧪 Think this could help someone else? [Star Math & Matter on Github](https://github.com/rajeevphysics/Obsidian-MathMatter) to help more learners discover it. ---