L'Hôpital's Rule (Calculus)

Given differentiable functions $f\left(x\right)$ and $g\left(x\right)$ over an interval which contains $a$, but may exclude it. Where ${\lim }_{x\to a}f\left(x\right)=0\vee \mathrm{\infty }$ and ${\lim }_{x\to a}g\left(x\right)=0\vee \mathrm{\infty }$^[1]:

$$\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$$

Example

Given the function $f\left(h\right)$:

$$\begin{array}{lrlr}& \underset{h\to 0}{\lim }f\left(h\right)& =\frac{3\ln (e+h)-3\ln \left(e\right)}{h}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{3\ln (e+h)-3}{h}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{0}{0}& \text{<span class="tml-eqn"></span>}\end{array}$$

Meaning we need to use L’Hôpital’s Rule, so the first thing we do is find the derivative of both the numerator and denominator:

$$\frac{d}{dx}[3\ln (e+h)-3]=\frac{3}{e+h}$$

$$\frac{d}{dx}\left[x\right]=1$$

So now we can find the limit of the function:

$$\begin{array}{lrlr}& \underset{h\to 0}{\lim }f\left(h\right)& =\frac{\frac{3}{e+h}}{1}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{3}{e+h}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{3}{e+0}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{3}{e}& \text{<span class="tml-eqn"></span>}\end{array}$$

1. L’Hôpital’s Rule by Gilbert Strang and Edwin “Jed” Herman via LibreTex Mathematics ↩