2601.sagews

L'Hopital's Rule for evaluating limits of quotients

 

Let's start with an example:  $\lim_{x \rightarrow 0} \frac{2x}{\sin(x)}$.

Call the numerator $f(x)$ and the denominator $g(x)$.  Both the numerator and denominator are zero when $x=0$, so we cannot evaluate this limit simply by substituting $x=0$.  We could try to evaluate the limit via a clever analysis, but this might be difficult.  L'Hopital's idea is to use local linear approximation.  Below is a graph of $g(x)=\sin(x)$ and its tangent line at $x=0$.  The tangent line is approximately the same as the graph of the function, when $x$ is near zero.

x=var('x')
g(x)=sin(x)
plot(g)
curveg=plot(g,(-1,2))
tang=plot(x, (-1,2),rgbcolor=(1,0,1))
show(tang+curveg)

Since $f(x)=2x$is linear, its tangent line is also$y=2x$.  Replacing the functions$f$and$g$with their tangent lines at$x=0$ we get:

for $x$ near zero $\frac{2x}{\sin(x)} \approx \frac{2x}{x}$, so $$\lim_{x \rightarrow 0} \frac{2x}{\sin(x)}=\lim_{x \rightarrow 0} \frac{2x}{x} = 2.$$

Theorem (L'Hopital's Rule):  If $f$ and $g$ are differentiable functions and $a$ is any real number or $+/- \infty$ and

• $\lim_{x\rightarrow a} f(x) = +/- \infty$ and $\lim_{x\rightarrow a} g(x) = +/- \infty$, or
• $\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} g(x) = 0$, then

$$\lim_{x\rightarrow a}\frac{f(x)}{g(x)} = \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}.$$

Let's try a few examples:

1.  $\lim_{x\rightarrow \infty} \frac{\ln(x)}{x}$

2.  $\lim_{x\rightarrow \infty} \frac{x^2}{2^x}$

3.  $\lim_{x\rightarrow 0} \frac{1-\cos(x)}{x^2}$