Question

• Prove L’Hopital’s rule.
• To proof: if \(f(a)=g(a)=0\), then \(\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f’(x)}{g’(x)}\)

Solution

There are multiple approaches to this

Prove from first principle - the way that always work

$$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f(x)-0}{g(x)-0}$$

\(\because f(a)=g(a)=0\)

$$ \begin{align*} \lim_{x\to a}\frac{f(x)}{g(x)}&=\lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}\\\\ &=\lim_{x\to a}\left(\frac{\space\space\dfrac{f(x)-f(a)}{x-a}\space\space}{\dfrac{g(x)-g(a)}{x-a}}\right)\\\\ &=\frac{\displaystyle\lim_{x\to a}\left(\dfrac{f(x)-f(a)}{x-a}\right)}{\displaystyle\lim_{x\to a}\left(\dfrac{g(x)-g(a)}{x-a}\right)} \end{align*} $$

By definition \(f’(a)=\displaystyle\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}\)

$$\therefore\quad\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f’(a)}{g’(a)}=\lim_{x\to a}\frac{f’(x)}{g’(x)}$$

It works, but it’s average, and not especially pleasant to look at. There’s a less generic way to do it, which in my opinion is really cool.

Linear approximation - quick and simple

$$f(x)\approx f(a)+f’(a)(x-a)$$

Linear approximation is accurate for values of \(x\) close to \(a\), and as \(x\to a\), the approximation becomes

$$f(x)=f(a)+f’(a)(x-a)$$

Therefore as \(x\to a\)

$$ \frac{f(x)}{g(x)}=\frac{f(a)+f’(a)(x-a)}{g(a)+g’(a)(x-a)} $$

We know \(f(a)=g(a)=0\)

$$ \begin{align*} &=\frac{f’(a)(x-a)}{g’(a)(x-a)}\\\\ &=\frac{f’(x)}{g’(x)} \end{align*} $$